My First Quartic Equation...

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Today we try to find the roots of the polynomial equation x^4+4x-1=0. We use some sneaky completion of the square as well as binomial theorems and the difference of two square to find our 4 roots out :) Enjoy! =D
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Want to know more about me? Watch my QnA! =D • Question and Answer Ti...

[@PapaFlammy69]

*_Hope you did like today's video

[@anshulrai677]

Nice I like this equation

[@shyeyebee]

I must say this is emoji is hideous lol

[@oni8337]

69

[@fromhl7619]

420

[@fuji_films]

@@oni8337 Pogger.

[@obinator9065]

Level 1 Thug: My first quadratic Level 9999 Euler: My first septic

[@luisbielmillan8467]

xd

[@dijkstra4678]

(271:20:51)

[@pierreabbat6157]

Septic tank: A seventh-degree algebraic surface containing sewage.

[@chubbybunny6272]

pathetic. i have had sepsis multiple times in my life.

[@That_One_Guy...]

My first septic tank : "It tastes shit"

[@DissonantSynth]

Never in my life I expected to find a mathematician calling himself "papa flammy" . This pleases me. More math memes!

[@nikhilnagaria2672]

Papa is a Chad enjenir wdym

[@TechToppers]

Ohh... We have a new guy in the gang.

[@nikhilnagaria2672]

@@TechToppers ?

[@ninu1510]

Josh j Did you go to school in bulgaria

[@DissonantSynth]

@@ninu1510 no???

[@pinklinecabs3967]

Flammable maths-Thanks for watching. Meanwhile me-Thanks for teaching.👍

[@chessdev5320]

I like how he pronounce 'y'

[@chixenlegjo]

Whai

[@hardikjoshi8557]

I have one girlfriend, mathematics. I date her everyday in night at my study table ordering some notebooks, pens, look at her beautiful face and she tells me everything about her, her matrices, her geometry, calculus. She asks me a bunch of new questions from whatever she explained and whenever i have failed she has motivated me. She is my only girlfriend. And credit goes to papa flammy for introducing me to her. Your Boi :^)

[@3rdDegree_]

[wipes tear from face] this is beautiful

[@kathanshah8305]

Maths is Futa

[@meinbuch9458]

I'm so deeply touched by this love story. Thank you for this masterpiece.

[@jakubabram9606]

plot twist: she is imaginary

[@kathanshah8305]

@@jakubabram9606 mine’s bigger

[@gamingstars8956]

this happens when real part of your teacher is zero

[@shubhmishra2158]

I think that you should have said, “this happens when imaginary part of your teacher is zero” because in that cause only will your teacher be real and this case will be considered special case because in general case it would be real part is zero, where we know that (real) = teacher like papa inflame. and (imaginary) = not like papa inflame.

[@gamingstars8956]

@@shubhmishra2158 i just thought real means which we see everyday (ordinary teachers) and immaginary is something special(papa flammy)

[@shubhmishra2158]

@@gamingstars8956 oh then it makes sense

[@assassin01620]

@@gamingstars8956 I thought you meant the teacher wasn't keeping it real.

[@Daniel-ef6gg]

This does demonstrate a means of solving a more general quartic equation. Try to manipulate, though change of variables and careful selection of coefficients, your equation into the form (ax^2+b)^2 - (cx+d)^2 = 0, and you can use more simple quadratic formula equations in order to finish the solution.

[@angelmendez-rivera351]

Yes, and this is actually equivalent to Cardano's method.

[@mba4677]

But what about the x cubed term?

[@angelmendez-rivera351]

@@mba4677 You can always transform a polynomial X^4 + A·X^3 + B·X^2 + C·X + D into a polynomial Y^4 + P·Y^2 + Q·Y + R via a Tschirnhaus transformation, namely by having Y = X + A/4. So the cubic monomial is not relevant. Every polynomial of degree n can be turned into a depressed polynomial of degree n with the transformation Y = X + A/n.

[@angelmendez-rivera351]

@@mba4677 Here is the explicit method. Suppose that the polynomial you want to transform is X^4 + 4·A·X^3 + B·X^2 + C·X + D. Every complex polynomial can be written in this form. Let X = Y - A. So X^4 + 4·A·X^3 + B·X^2 + C·X + D = (Y - A)^4 + 4·A·(Y - A)^3 + B·(Y - A)^2 + C·(Y - A) + D = (Y^4 - 4·A·Y^3 + 6·A^2·Y^2 - 4·A^3·Y + A^4) + 4·A·(Y^3 - 3·A·Y^2 + 3·A^2·Y - A^3) + B·(Y^2 - 2·A·Y + A^2) + C·(Y - A) + D = Y^4 - (6·A^2 - B)·Y^2 + (8·A^3 - 2·A·B + C)·Y - (3·A^4 - A^2·B + A·C - D). Let P = -(6·A^2 - B), Q = 8·A^3 - 2·A·B + C, R = -(3·A^4 - A^2·B + A·C - D). This completes the transformation. Once you have the polynomial Y^4 + P·Y^2 + Q·Y + R, you can apply the method from the comment to factor the polynomial.

[@Daniel-ef6gg]

Note that there will be a cubic term involved in the substitution, so solving a quartic in this method does require solving a characteristic cubic in general.

[@chemistryhorseman2296]

On of the best math channel on KZbin)

[@PapaFlammy69]

:)

[@tac0cat14]

You mean just the best?

[@chemistryhorseman2296]

@@tac0cat14 of course

[@d.bcooper7691]

*best

[@leif1075]

@@PapaFlammy69 Why did you solve it this way? Hope you can respond.

[@RC32Smiths01]

Congrats on pulling this one off man!

[@fuji_films]

2:20 x plus woah.

[@ndaktau67q]

You •_• #saveblue

[@goodplacetostop2973]

16:43

[@user-zu6ts5fb6g]

i jumped

[@orenfivel6247]

H.W de-nest the square-roots and rationalize the denominators

[@davidvarga5298]

cool and good channel and videos, helped me find something good to do with my time, keep it going

[@PapaFlammy69]

:)

[@angelmendez-rivera351]

You cited the Abel-Ruffini theorem, but I think you actually meant to cite the fundamental theorem of algebra. The fundamental theorem of algebra states that the field of complex numbers is algebraically closed, which is to say that every polynomial with complex coefficients can be decomposed into a number of monomial factors equal to its degree.

[@MrBeen992]

No, he mentioned Abel-Ruffini to state that the a fourth degree polynomial equation (with rational coefficients) can be solved in radicals. Thats not implied by the FTA.

[@angelmendez-rivera351]

@@MrBeen992 No, that is not what he said in the video. Watch the video again. Here is the timestamp, 1:04 - 1:32. He definitely did just say, "by the Abel-Ruffini theorem, all four roots to the polynomial exist, if the field is C or R." He never mentioned radical expressions, not even once. Next time, I suggest you actually pay close attention before trying to correct me.

[@MrBeen992]

@@angelmendez-rivera351 Sir, with all due respect you stopped listening at 1:30. Just listen at 1:30 how he explicitly say "If its fifth degree you cant express all the roots in radicals, but in fourth you can. Thanks Abel-Ruffini". FTA was already known before Abel-Ruffini. And it doesnt make sense for him to try to solve algebraically a 5th degree or higher polynomial that he doesnt know if it can be solved by radicals.

[@angelmendez-rivera351]

@@MrBeen992 *Sir, with all due respect you stopped listening at **1:30**. Just listen at **1:30** how he explicitly say "If its fifth degree you cant express all the roots in radicals, but in fourth you can. Thanks Abel-Ruffini".* Yes, I know he said this. This does not disprove the fact that I have already pointed out: that he used the Abel-Ruffini theorem to justify the claim that the four roots of a quartic polynomials exist over C, and not to say that they are expressible in terms of radicals. Thus, *you are wrong.* Just accept that you made a mistake, and move on. You said he did not do this, and I showed you the timestamp where he did say it. Whatever he said afterwards changes absolutely nothing. *FTA was already known before Abel-Ruffini.* I know this. This is irrelevant. *And it doesnt make sense for him to try to solve algebraically a 5th degree or higher polynomial that he doesnt know if it can be solved by radicals.* This is the stupidest reply I have heard in a while. Go to the KZbin search bar and type "Solving a quintic equation." You will find at least a dozen of results. If he wanted to make a video solving a quintic polynomial equation, then he could do it, and in that scenario, the Abel-Ruffini theorem would simply be irrelevant. It is irrelevant in this video too. The only thing that is required to solve an equation is that the equation be solvable, and that solvability is given by the FTA, which is what Jensen was referring to, not the Abel-Ruffini theorem, which is irrelevant, since Jensen was not concerned with radical expressions, he was concerned with whether the polynomial was guaranteed to have 4 roots or not in the context of complex numbers or real numbers.

[@MrBeen992]

@@angelmendez-rivera351 Sir, I dont know why you are so upset. When I mentioned solving a quintic I refer explicitly to solving in radicals. As you know this is not possible, in general. Thanks Abel-Ruffini. He is solving a quartic equation he knows has a radicals solution. As to the numbers of solutions and the field of solvability...thats already implied with Abel-Ruffini. So, if it satisfies you, you are completely right. But in a narrow sense. Have a good day, sir.

[@bendakstarkiller603]

I like your funny words magic man. I haven't done math stuff in a long time, but you were in my recommendations for some reason and I wish that I had these videos in Highschool.

[@andreiiuliantoma6303]

All my respect Pappa Flammy!

[@lucakoch3432]

Hey papa flammy, could you (or someone else) please explain me why you sometimes use an equal sign with 3 lines and what it is supposed to mean. That would be great ^^

[@noneofyourbeeswax8045]

It means "is identical to" and its use is a bit weird there. You might use it when you have a proof and you see okay that bit I proved here and this bit I proved here are indetical, then you'd use three lines. In the case of polynomial =0 you should rather use an exclamation mark on top of it to express that in this case it is supposed to be 0

[@user-rv9vk8by5i]

Usually, it means "identity", or "always/strictly equal to" When solving an equation, you'll use the regular equals sign. When presenting a fact, you can either say that the left hand side equals the right hand side, for all values of x Or you can simply use the triple line equals, to show that it is an identity That said, I have no clue why flammy used the identity sign - there's only 4 values of x that satisfy the equation, it isn't true for all x

[@angelmendez-rivera351]

@@user-rv9vk8by5i Yes, Papa Flammy's usage of the symbol is just incorrect, in this case.

[@PapaFlammy69]

Ye, sometimes I just use it 'cause it looks cooler lol. Sorry for the confusion ^^'

[@lucakoch3432]

@@PapaFlammy69 Haha okay 😂👌

[@kraze4kicks822]

Easy, just derive the quartic formula and plug in

[@ricardoparada5375]

I would simply use the quartic formula but that’s probably cuz I’m built different lmao

[@JacoTheDeadRuler]

Wow! Papa Flammys almost at 1/4 of a million subs :D

[@PapaFlammy69]

ayyyyyy :D

[@ndaktau67q]

•_• b r u h #saveblue

[@kathanshah8305]

What will you do with your figurines doujinshi when brother/ sister flammy comes ?

[@reclaimer2019]

Wait y'all do equations? I just change the number of "x" till i get the result.

[@ElectroNeutrino]

You can also pull out 1/sqrt(2) from the radicals on the solutions to give x12 = (1 +- i * sqrt( 2 * sqrt(2) + 1)) / sqrt(2), x34 = (1 +- sqrt( 2 * sqrt(2) - 1)) / sqrt(2). That way you avoid nested radicals. And if you want a single expression that covers all 4 solutions, you can write x = (1 +- sqrt( +- 2 * sqrt(2) - 1)) / sqrt(2).

[@raphaelteng2344]

Underrated comment

[@mptness4389]

5:38 The chalkboard has grown a face that says it has become annoyed.

[@muskyoxes]

I was trying to see which coefficients this procedure works for and it seemed to me that, for x^4 + bx + c = 0, you need to solve for d in d^3 - cd = (b^2)/8, either by guessing or hard work. that value of d then becomes the coefficient of x^2 needed for completing the square, so in effect the quartic is reduced to a cubic.

[@mudkip_btw]

Just solve it numerically papa

[@adityaekbote8498]

Feeling proud Papa

[@PapaFlammy69]

:)

[@themightyripples6582]

Im just starting with maths (first year at uni, going into second). You said a polynomial over fields C or R can have roots, of course! But, what kind of fields are there where a polynomial can't have any roots? I've had a short introduction in linear algebra to defining fields (like, where you define the rules of addition and multiplication, etc) but its hard for me to picture immediately what a field where there are no roots of polynomials, Particularly odd degree polynomials. It'd be interesting to find out though!

[@frei6833]

Im not an algebra expert but I did Algebra I and II at Uni, so I might help. Answering your question..the polinomial p(T)=T²+1 does not have any roots in R[T], but it has 2 roots (i and -i) in C[T]. So as you can see R[T] is a thing where you can find many polinomials without roots. Also Q and finite fields are interesting to see. In the case your goal was to find a field F such that no polinomial is solvable in F[T]...well I don't think there is. It should be a field where even linear polynomials like p(T)=T or p(T)=T-1 couldn't be solvable. This means that F would be an empty space, which, as far as I am concerned, is not even a field.

[@dan-valentin3023]

I love it when he says "y". Wha ? 😂😂😂

[@asonjarossa2023]

I wonder if you could use this trick for all even orders; but how would we do it with odd orders? I remember a formula for simplifying a cubed function (can't remember it on the top of my head right now), but I wonder if we could do the same technique here, combine it with the original postulate I described earlier, would that help in simplifying our equations even OR odd?

[@nikhilnagaria2672]

0:05 that meme tho 😂

[@syedinayat3548]

vater flamy why did you turn red while doing these equation(did andrew had a barbeque party(・o・).

[@PapaFlammy69]

yeye

[@vedants.vispute77]

Really good the video is

[@desertrainfrog1691]

This makes me remember something I saw recently about finding/expressing the roots of a polynomial equation in terms of trig functions. For example, x^4+x^3+x^2+x+1 can be factored as (x^2-2xcos(2π/5)+1)(x^2-2xcos(4π/5)+1) What the heccin? I don't know that I've seen this idea used on the channel yet, but I'd be curious to see you do something relating to it.

[@raphaelteng2344]

My first approach was to sub in and find roots (eg x=1, x=-1, x=2, x=-2, x=1/2...) & give up In my defence, it works most of the time

[@kathanshah8305]

So busy wood working haven’t even ripped off tags from the new clothes

[@PapaFlammy69]

yeye 3:

[@olopadolf8156]

SD CARD EXTRAVAGANZA is the best moment of this video !

[@FisicaeMatematica]

Brillant !!!!!!

[@teacheryiespeseth162]

Thanks you ❤️ teacher ✅

[@theevilmathematician]

I use Brilliant and it makes me Brilliant and helps dis Brilliant channel.

[@PapaFlammy69]

[@sharkknight618]

2:22 what is waa

[@PapaFlammy69]

y

[@jareddiscipio1768]

@@PapaFlammy69 no... it’s actually wa

[@devanshisharma2447]

i had the same eqn in a ques in a test i gave today magic

[@PapaFlammy69]

nice =D

[@kathanshah8305]

Are you posting to 3B1B cause you are a prime target You are teacher and KZbinr

[@PapaFlammy69]

what?

[@kathanshah8305]

@@PapaFlammy69 watch the new 3 blue one brown video

[@bibaswanroy2161]

Can u do some on Reimann Hypothesis, Lord Papa Flammy

[@noneofyourbeeswax8045]

The way you say y drives me absolutely crazy and not in a good way

[@PapaFlammy69]

:D

[@poutineausyropderable7108]

Physicist method: x^4 + 4x = 1 Use perturbation. Drop a part. Which one? x^4 + 4x = 1 + err(y) Drop x^4 : x= 1/4. Error in y is (1/4)^4. drop 4x : x =1. Error is in y 4. Better drop x^4. 4x = 1 - x^4 x = 1 - x^4 / 4 X0 = 0 x_n+1 = 1/4 - (x_n)^4 /4 x1 = 1/4. X2< x1. x3> x2. X infinity is between x1 and x2. X infinity is between x2 and x3. X infinity is between Xn and Xn+1. Just pick your Xn. The gap size probably gets smaller exponentially. Or you know, just use newtons method.

[@pi4313]

11:46 where did you get the over 2? Shouldn't it just be root 2?

[@goblinkoma]

When you use the quadratic formula you may aswell use the quartic formula.. jk this is much cooler

[@PapaFlammy69]

:p

[@fuji_films]

I like the 69 formula.

[@emrecanaslan8699]

@@fuji_films nice.

[@kasmirperriman9360]

"Wah" -papa "the beast" flammy

[@garvett6660]

These fettucine are too good to remain untold of

[@danielesantospirito5743]

No, no, he's got a point

[@PapaFlammy69]

yeye

[@Zavstar]

How did you decide to choose to add and subtract 1 we could have chosen 4 or 9 too

[@PapaFlammy69]

yup! But it wouldn't factor nicely :)

[@Zavstar]

@@PapaFlammy69 where can i buy this intelligence

[@lewisfoskett122]

It doesnt matter because you subtract it off anyway, 1 is just easier to work with

[@angelmendez-rivera351]

Half of 2 is 1, and 1 squared is 1. Hence he chose 1.

[@user-fi6if8gx3g]

This really tells alot about the industrial society

[@ndaktau67q]

•_• #saveblue

[@asyauzun6605]

hii papa

[@PapaFlammy69]

Hey Asya

[@user-sj2vg8hb5q]

How about combining math with donations for the floods?

[@PapaFlammy69]

what flood?

[@boldizsarmann5023]

Probably the one in the west of Germany

[@user-sj2vg8hb5q]

@@boldizsarmann5023 yes.

[@ethanbeck5535]

Very cool non bounded integral, integral of e^-x^(2abs(a)) where a=\=0

[@AbhishekSinha-mo7oo]

Heyyyy Papa love from India...can you please do some videos on Quadrics like cylinder...🙏🏻

[@user-ry6ey8gq3t]

I love Brilliant 🤩

[@PapaFlammy69]

same!

[@kathanshah8305]

@@PapaFlammy69 come one you had perfect chance to say brilliant

[@kathanshah8305]

Brilliant

[@bhargav7215]

Initially Watching a dot on Papa's forehead I've thought he has arrived from a temple 😁

[@niltondasilva1645]

Daria para resolver a equação do quarto grau x^4+ 9x^1-1= 0 usando o mesmo caminho que você resolveu essa x4+4x^1-1 = 0 ?

[@Astromath]

But if you use the quadratic formula, don't you have to divide sqrt(b²-4ac) also by 2a??

[@angelmendez-rivera351]

Not necessarily. The formula he used is -b/(2·a) +/- sqrt([-b/(2·a)]^2 - c), which is equivalent to the typical quadratic formula.

[@wadrianalefe6490]

@@angelmendez-rivera351 I tried expanding that in here and got to --b/(2a) +/-- [1/(2a)]*sqrt[(-b)^2- 4a^2*c]. Should'nt that be -b/(2a)+/- [1/(2a)]*sqrt((-b)^2-4ac) instead? Did I do something wrong?

[@wadrianalefe6490]

I don't know why but youtube is risking some terms in my comment. It is supposed to be minus b over 2a +/-

[@shreysingh9327]

He did..He used the above mentioned formula which is actually similar to the one you wrote..

[@insimplifiable920]

when the symbol at the start isnt an equal sign but represents mod congruence:

[@RodrigoMartinelliMatematica]

Quintic equation has been cracked!

[@neilgerace355]

If you rub caustic soda on your back, you will feel even more basic. Briefly.

[@PapaFlammy69]

xD

[@niccologeraci9175]

>mfw no quartic formula

[@PapaFlammy69]

r i p

[@78anurag]

Nooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo

[@SHUBHAMGI]

I hate it when people put square bracket inside round bracket, nice video

[@PapaFlammy69]

lol

[@fromhl7619]

X = wa X² = wawa

[@davidbrisbane7206]

Virgin quartic equation ... I am blushing papa.

[@perveilov]

My dumbass brain: "ouai"

[@urisinger3412]

i did this in my head

[@jedmaegraith417]

2x + wha

[@dqrksun]

3:10

[@phyarth8082]

reduction biquadratic equation into quadratic :)

[@PapaFlammy69]

yas :3

[@sharpfang]

This is one part which I really hate about maths. We're nowhere near a solution so let's add and subtract 2x^2. Sure, you *can*. You can also add and subtract e^pi, or cos(x). What is missing is how do you choose what to add and subtract - how do you know a'priori this will lead to a solution.

[@Rakesh37187]

That pretty much comes down to experience. If you see that hey by doing something it's much easier because you've seen/done it before or done a problem that's similar to it before you'll obviously use that again.

[@sharpfang]

@@Rakesh37187 Yeah. Experience, intuition, hunch, vague similarity to prior cases. What is this, dowsing? astrology? School exercises are at least grounded in faith in sanity of the author of the exercise. Real-life calculations? Spend seven pages on ever-growing equation, then finally manage to reduce it until you arrive upon the ground-breaking discovery that 0=0.

[@ganimede17]

I agree with you: when someone has "the curse of knowledge" (i am an expert in a field but I have difficulties in communicating what I do to non-experts) and does these arbitrary manipulations, i always search for a motivation beyond "it works". For this equation i would think x**4+4x-1 as the product of 2 quadratic polynomials x**2 + ax + b and x**2 + cx + d with a,b,c,d real numbers. Such polynomials must exist for the fundamental theorem of algebra. Then i would compute a,b,c,d with the definition of identity for polynomials.

[@aiseop31415]

Recommend best mathematics books for self study...... 😀😀

[@MINEXKILLER]

Actually the lighting is better after the SD extravaganza also keep the shit memes coming

[@ndaktau67q]

You kill Among US blue vote!!! You!!! 😂😂#saveblue

[@Dude-cf8hb]

The way you say "y" is so cute.

[@PapaFlammy69]

:D

[@Dude-cf8hb]

@@PapaFlammy69 Also I don't understand any of this but it's OK I'm tryna learn from you.

[@aiseop31415]

Is it "y" or "wa" Anyways nice explanation

[@enriqueniemannconcha7267]

Yeye

[@mateozenteno4566]

Hallo Papa, ich komme im September nach Deutschland, um mit meinem Studium anzufangen. Danke für deine tolle Videos. Grüße aus Lateinamerika :D

[@nachiketakumar9645]

I think its German???

[@PapaFlammy69]

[@nachiketakumar9645]

@@PapaFlammy69 papa flammy, please can you make videos on multidimension topology 🙏

[@fuji_films]

@@nachiketakumar9645 Imagine asking twice.

[@nachiketakumar9645]

@@fuji_films but what?

[@thegodsofai3845]

Lmao what happened to the audio 1:11

[@PapaFlammy69]

My mic is doing weird shit these days and I have no idea why 3:

[@sebastianschubert7945]

Okay this channel is pretty good but I don't understand where the weird pronunciation of "y" is coming from? Please explain 😊

[@tl1989]

Ciao? Lol!

[@danipent3550]

Does the legend reply?

[@PapaFlammy69]

yeye

[@danipent3550]

@@PapaFlammy69 continue these vids man 🔥 🙌 🙌 🔥 🔥

[@meinegute4261]

"Anal 1 für Informatiker"

[@PapaFlammy69]

yeye

[@pardeepgarg2640]

17cmnt 781 viewer aka 71×11 A prime viewer :)

[@fuji_films]

17 centimeters? 😭

[@pardeepgarg2640]

@@fuji_films Huh

[@bishwayanroychowdhury]

What is wa?

[@PapaFlammy69]

y

[@bishwayanroychowdhury]

@@PapaFlammy69 ok

[@KatzRool]

wah

[@PapaFlammy69]

wuahhhhh

[@pandabearguy1]

Now do your first quadratic equation

[@PapaFlammy69]

ez

[@asmocak1053]

Can we solve it by treating "2" as a variable and use the quadratic formula with respect to "2" ?

[@CFneo98]

i don't think so, 2 is a constant by definition, you cannot solve for 2 because 2=2

[@Ultiminati]

@@CFneo98 Did you not see "How real men solve equations" video :D The most viewed thing on this channel and i also was introduced to this channel from that video :)

[@user-ox5ml5ee9v]

@@CFneo98 this was a joke i think

[@oxil3473]

Wierd

[@user-us4ws9px2s]

are you a member of MENSA?

[@PapaFlammy69]

hell no lol

[@user-us4ws9px2s]

@@PapaFlammy69 maybe you should try to be member? Is it always a good idea to factor polynomials with grade 4 without power 3 and 2 with completing square?

[@cooldawg2009]

Why does his quadratic formula looks so weird? Isn’t it supposed to be (-b (+/-) sqrt(b^2-4ac))/2a?

[@calvinjackson8110]

Tricky. If you are not clever, you may as well throw in the towel. I would much rather have a method I can rely on instead of hoping to be clever to pull a rabbit out of a hat.

[@user-ud6xw3js3r]

idgaf i got my op ruffini theorem it runs through anything, from 3th to any grade equations, that bastard eats a lot of time tho

[@nachiketakumar9645]

Dad, please can you make videos on multidimension topology 🙏

[@user-iz6gi1rf4t]

Ferrari is so ferrari...

[@David-km2ie]

Or you use the ferrari formula

[@idudheebsbzdudbdhddh]

im so fucking bad at maths why am i watching this

[@PapaFlammy69]

:D

[@vishweshbhalerao6798]

Cicada 3301

[@ndaktau67q]

I don't know mathematics high level and autoomous equation 👍 #saveblue

[@TheScienceGuy10]

Wha

[@enejidjsi5939]

wa

[@PapaFlammy69]

wuahhhh