Who found the proof for the quartic formula?

Question

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.

Answer 1

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:

• Descartes’ Method for Constructing Roots of Polynomials with ‘Simple’ Curves - 'Depressed' Quartics and Cubics
• Descartes method (fourth degree equation)

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).