<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] What has not been mentioned thus far is that one can in fact use any number of "auxiliary cubics" in the solution of the quartic equation. Don Herbison-Evans, in this page (Wayback link, as the original page has gone kaput), which was adapted from his technical report, mention five such possible auxiliary cubics.
Given the quartic equation
$$x^4 + ax^3 + bx^2 + cx + d = 0$$
the five possible auxiliary cubics are referred to in the document as
Christianson-Brown:
$$y^3 +\frac{4a^2b - 4b^2 - 4ac + 16d - \frac34a^4}{a^3 - 4ab + 8c}y^2 + \left(\frac3{16}a^2 - \frac{b}{2}\right)y - \frac{1}{64}(a^3 + 4a b - 8c) = 0$$
Descartes-Euler-Cardano:
$$y^3 + \left(2b - \frac34 a^2\right)y^2 + \left(\frac3{16}a^4 - a^2b + ac + b^2 - 4d\right)y - \frac{1}{64}(a^3 + 4a b - 8c)^2 = 0$$
Ferrari-Lagrange
$$y^3 + by^2 + (ac - 4d)y + a^2d + c^2 - 4bd = 0$$
Neumark
$$y^3 - 2by^2 + (ac + b^2 - 4d)y + a^2d - abc + c^2 = 0$$
Yacoub-Fraidenraich-Brown
$$(a^3 - 4ab + 8c)y^3 + (a^2b - 4b^2 + 2ac + 16d)y^2 + (a^2c - 4bc + 8ad)y + a^2d - c^2 = 0 $$
See the page for how to obtain the quadratics that will yield the solutions to the original quartic equation from a root of the auxiliary cubic.
Let me also mention this old ACM algorithm in Algol. Netlib has a C implementation of that algorithm.
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