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?

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.
