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A quartic equation is a 4th degree polynomial, in the form of $ax^4+bx^3+cx^2+dx+e$. There are 4 different formulae for the 4 roots of the quartic equation. Here are the formulae: $$x_1=-\frac{b}{4 a}-\frac{1}{2} \sqrt{\frac{\sqrt[3]{2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e+\sqrt{\left(2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e\right)^2-4 \left(c^2-3 b d+12 a e\right)^3}}}{3 \sqrt[3]{2} a}+\frac{b^2}{4 a^2}-\frac{2 c}{3 a}+\frac{\sqrt[3]{2} \left(c^2-3 b d+12 a e\right)}{3 a \sqrt[3]{2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e+\sqrt{\left(2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e\right)^2-4 \left(c^2-3 b d+12 a e\right)^3}}}}-\frac{1}{2} \sqrt{-\frac{\sqrt[3]{2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e+\sqrt{\left(2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e\right)^2-4 \left(c^2-3 b d+12 a e\right)^3}}}{3 \sqrt[3]{2} a}+\frac{b^2}{2 a^2}-\frac{4 c}{3 a}-\frac{\sqrt[3]{2} \left(c^2-3 b d+12 a e\right)}{3 a \sqrt[3]{2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e+\sqrt{\left(2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e\right)^2-4 \left(c^2-3 b d+12 a e\right)^3}}}-\frac{-\frac{b^3}{a^3}+\frac{4 c b}{a^2}-\frac{8 d}{a}}{4 \sqrt{\frac{\sqrt[3]{2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e+\sqrt{\left(2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e\right)^2-4 \left(c^2-3 b d+12 a e\right)^3}}}{3 \sqrt[3]{2} a}+\frac{b^2}{4 a^2}-\frac{2 c}{3 a}+\frac{\sqrt[3]{2} \left(c^2-3 b d+12 a e\right)}{3 a \sqrt[3]{2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e+\sqrt{\left(2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e\right)^2-4 \left(c^2-3 b d+12 a e\right)^3}}}}}}$$

$$x_2=-\frac{b}{4 a}-\frac{1}{2} \sqrt{\frac{\sqrt[3]{2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e+\sqrt{\left(2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e\right)^2-4 \left(c^2-3 b d+12 a e\right)^3}}}{3 \sqrt[3]{2} a}+\frac{b^2}{4 a^2}-\frac{2 c}{3 a}+\frac{\sqrt[3]{2} \left(c^2-3 b d+12 a e\right)}{3 a \sqrt[3]{2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e+\sqrt{\left(2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e\right)^2-4 \left(c^2-3 b d+12 a e\right)^3}}}}+\frac{1}{2} \sqrt{-\frac{\sqrt[3]{2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e+\sqrt{\left(2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e\right)^2-4 \left(c^2-3 b d+12 a e\right)^3}}}{3 \sqrt[3]{2} a}+\frac{b^2}{2 a^2}-\frac{4 c}{3 a}-\frac{\sqrt[3]{2} \left(c^2-3 b d+12 a e\right)}{3 a \sqrt[3]{2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e+\sqrt{\left(2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e\right)^2-4 \left(c^2-3 b d+12 a e\right)^3}}}-\frac{-\frac{b^3}{a^3}+\frac{4 c b}{a^2}-\frac{8 d}{a}}{4 \sqrt{\frac{\sqrt[3]{2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e+\sqrt{\left(2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e\right)^2-4 \left(c^2-3 b d+12 a e\right)^3}}}{3 \sqrt[3]{2} a}+\frac{b^2}{4 a^2}-\frac{2 c}{3 a}+\frac{\sqrt[3]{2} \left(c^2-3 b d+12 a e\right)}{3 a \sqrt[3]{2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e+\sqrt{\left(2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e\right)^2-4 \left(c^2-3 b d+12 a e\right)^3}}}}}}$$

$$x_3=-\frac{b}{4 a}+\frac{1}{2} \sqrt{\frac{\sqrt[3]{2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e+\sqrt{\left(2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e\right)^2-4 \left(c^2-3 b d+12 a e\right)^3}}}{3 \sqrt[3]{2} a}+\frac{b^2}{4 a^2}-\frac{2 c}{3 a}+\frac{\sqrt[3]{2} \left(c^2-3 b d+12 a e\right)}{3 a \sqrt[3]{2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e+\sqrt{\left(2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e\right)^2-4 \left(c^2-3 b d+12 a e\right)^3}}}}-\frac{1}{2} \sqrt{-\frac{\sqrt[3]{2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e+\sqrt{\left(2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e\right)^2-4 \left(c^2-3 b d+12 a e\right)^3}}}{3 \sqrt[3]{2} a}+\frac{b^2}{2 a^2}-\frac{4 c}{3 a}-\frac{\sqrt[3]{2} \left(c^2-3 b d+12 a e\right)}{3 a \sqrt[3]{2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e+\sqrt{\left(2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e\right)^2-4 \left(c^2-3 b d+12 a e\right)^3}}}+\frac{-\frac{b^3}{a^3}+\frac{4 c b}{a^2}-\frac{8 d}{a}}{4 \sqrt{\frac{\sqrt[3]{2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e+\sqrt{\left(2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e\right)^2-4 \left(c^2-3 b d+12 a e\right)^3}}}{3 \sqrt[3]{2} a}+\frac{b^2}{4 a^2}-\frac{2 c}{3 a}+\frac{\sqrt[3]{2} \left(c^2-3 b d+12 a e\right)}{3 a \sqrt[3]{2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e+\sqrt{\left(2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e\right)^2-4 \left(c^2-3 b d+12 a e\right)^3}}}}}}$$

$$x_4=-\frac{b}{4 a}+\frac{1}{2} \sqrt{\frac{\sqrt[3]{2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e+\sqrt{\left(2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e\right)^2-4 \left(c^2-3 b d+12 a e\right)^3}}}{3 \sqrt[3]{2} a}+\frac{b^2}{4 a^2}-\frac{2 c}{3 a}+\frac{\sqrt[3]{2} \left(c^2-3 b d+12 a e\right)}{3 a \sqrt[3]{2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e+\sqrt{\left(2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e\right)^2-4 \left(c^2-3 b d+12 a e\right)^3}}}}+\frac{1}{2} \sqrt{-\frac{\sqrt[3]{2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e+\sqrt{\left(2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e\right)^2-4 \left(c^2-3 b d+12 a e\right)^3}}}{3 \sqrt[3]{2} a}+\frac{b^2}{2 a^2}-\frac{4 c}{3 a}-\frac{\sqrt[3]{2} \left(c^2-3 b d+12 a e\right)}{3 a \sqrt[3]{2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e+\sqrt{\left(2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e\right)^2-4 \left(c^2-3 b d+12 a e\right)^3}}}+\frac{-\frac{b^3}{a^3}+\frac{4 c b}{a^2}-\frac{8 d}{a}}{4 \sqrt{\frac{\sqrt[3]{2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e+\sqrt{\left(2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e\right)^2-4 \left(c^2-3 b d+12 a e\right)^3}}}{3 \sqrt[3]{2} a}+\frac{b^2}{4 a^2}-\frac{2 c}{3 a}+\frac{\sqrt[3]{2} \left(c^2-3 b d+12 a e\right)}{3 a \sqrt[3]{2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e+\sqrt{\left(2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e\right)^2-4 \left(c^2-3 b d+12 a e\right)^3}}}}}}$$
Yep, they are really long, I apologize for the loading time. I want to know who spent so much time finding these formulae and how they found them.

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    $\begingroup$ In this video, Mathologer shows how one solves the cubic equation. Then at the very end how you use that to solve the quartic in just a few minutes. Have a look. Spoiler: No one found that formula. They simplified the problem, solved the simplified problem, then transformed back. Note how many terms and expressions in your formulas are basically the same, only with a sign change here and there. That's the result of loads of simplifying substitutions. $\endgroup$
    – Arthur
    Dec 17, 2020 at 6:46
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    $\begingroup$ en.m.wikipedia.org/wiki/Quartic_equation $\endgroup$ Dec 17, 2020 at 8:13
  • $\begingroup$ Hopefully this will be migrated to hsm.se. $\endgroup$
    – J.G.
    Dec 13, 2021 at 7:09

1 Answer 1

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Many years ago (in the context of "Jugend forscht"), I retraced the development of these really old formulas (from degree 3 up to some special cases of degree 5) that have their origin in the 16th century. On page 15 in the document "Jugend Forscht" (PDF) you find the case of degree 4.

As I remember, René Descartes developed a method for calculating the four roots of an quartic equation. The following links might be helpful:

Ron Irving's book "Beyond the Quadratic Formula" gives a nice chapter on Descartes’ method (page 149) and on Ferrari's method (page 146), who was apparently the first discoverer of these formulas.

Edit:

As correctly pointed out by the comment of WhatsUp, Ferrari discovered first, how to find the roots of a quartic equation. Cardano published it originally in his book "Ars Magna" (PDF in Latin language). A nice essay explaining it in English can be found in the chapter "Cardano’s Ars Magna" (PDF).

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  • $\begingroup$ In this wiki section, it is said that the solution is found by Lodovico Ferrari in 1540 and published by Gerolamo Cardano in 1545. $\endgroup$
    – WhatsUp
    Dec 13, 2021 at 1:10
  • $\begingroup$ Ok - that is very intersting. I knew that Ferrari (and later Cardano) discovered the formula for solving cubic equations (called Cardano Formula). I will check this carefully and fix my answer. $\endgroup$ Dec 13, 2021 at 1:14

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