<s>[INST] Is there a general formula for solving Quartic (Degree $4$) equations? There is a general formula for solving quadratic equations, namely the Quadratic Formula, or the Sridharacharya Formula:

$$x = \frac{ -b \pm \sqrt{ b^2 - 4ac } }{ 2a } $$

For cubic equations of the form $ax^3+bx^2+cx+d=0$, there is a set of three equations, one for each root.
Is there a general formula for solving equations of the following form [Quartic Equations]?
$$ ax^4 + bx^3 + cx^2 + dx + e = 0 $$
How about for higher degrees? If not, why not?
 [/INST] We can reduce the problem of solving the general quartic to merely solving a quadratic. Given,
$$x^4+ax^3+bx^2+cx+d=0$$
Then the four roots are,
$$x_{1,2} = -\frac{a}{4}+\frac{\color{red}\pm\sqrt{u}}{2}\color{blue}+\frac{1}{4}\sqrt{3a^2-8b-4u+\frac{-a^3+4ab-8c}{\color{red}\pm\sqrt{u}}}\tag1$$
$$x_{3,4} = -\frac{a}{4}+\frac{\color{red}\pm\sqrt{u}}{2}\color{blue}-\frac{1}{4}\sqrt{3a^2-8b-4u+\frac{-a^3+4ab-8c}{\color{red}\pm\sqrt{u}}}\tag2$$
where,
$$u = \frac{a^2}{4}-\frac{2b}{3} +\frac{1}{3}\left(v_1^{1/3}\zeta_3+\frac{b^2 - 3 a c + 12 d}{v_1^{1/3}\zeta_3}\right)\tag{3a}$$
or alternatively,
$$u = \frac{a^2}{4}-\frac{2b}{3} +\frac{1}{3}\left(\color{blue}{v_1^{1/3}+v_2^{1/3}}\right)\tag{3b}$$
with $v_1$ any non-zero root of the quadratic,
$$v^2 + (-2 b^3 + 9 a b c - 27 c^2 - 27 a^2 d + 72 b d)v + (b^2 - 3 a c + 12 d)^3 = 0$$
and a chosen cube root of unity $\zeta_3^3 = 1$ such that $u$ is also non-zero. (Normally, use $\zeta_3=1$, but not when $a^3-4ab+8c = 0$.)
P.S. This is essentially the method used by Mathematica, though much simplified for aesthetics.
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