Is there a general formula for solving 4th degree equations (quartic)? There is a general formula for solving quadratic equations, namely the Quadratic Formula:

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

For third degree 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?
$$ ax^4 + bx^3 + cx^2 + dx + e = 0 $$
How about for higher degrees? If not, why not?
There is, in fact, a general formula for solving quartic (4th degree polynomial) equations.  As the cubic formula is significantly more complex than the quadratic formula, the quartic formula is significantly more complex than the cubic formula.  Wikipedia's article on quartic functions has a lengthy process by which to get the solutions, but does not give an explicit formula.

Beware that in the cubic and quartic formulas, depending on how the formula is expressed, the correctness of the answers likely depends on a particular choice of definition of principal roots for nonreal complex numbers and there are two different ways to define such a principal root.

There cannot be explicit algebraic formulas for the general solutions to higher-degree polynomials, but proving this requires mathematics beyond precalculus (it is typically proved with Galois Theory now, though it was originally proved with other methods).  This fact is known as the Abel-Ruffini theorem.

Also of note, Wolfram sells a poster that discusses the solvability of polynomial equations, focusing particularly on techniques to solve a quintic (5th degree polynomial) equation.  This poster gives explicit formulas for the solutions to quadratic, cubic, and quartic equations.

edit:  I believe that the formula given below gives the correct solutions for x to $ax^4+bx^3+cx^2+d+e=0$ for all complex a, b, c, d, and e, under the assumption that $w=\sqrt{z}$ is the complex number such that $w^2=z$ and $\arg(w)\in(-\frac{\pi}{2},\frac{\pi}{2}]$ and $w=\sqrt[3]{z}$ is the complex number such that $w^3=z$ and $\arg(w)\in(-\frac{\pi}{3},\frac{\pi}{3}]$ (these are typically how computer algebra systems and calculators define the principal roots).  Some intermediate parameters $p_k$ are defined to keep the formula simple and to help in keeping the choices of roots consistent.

Let: \begin{align*}
p_1&amp;=2c^3-9bcd+27ad^2+27b^2e-72ace
\\\\
p_2&amp;=p_1+\sqrt{-4(c^2-3bd+12ae)^3+p_1^2}
\\\\
p_3&amp;=\frac{c^2-3bd+12ae}{3a\sqrt[3]{\frac{p_2}{2}}}+\frac{\sqrt[3]{\frac{p_2}{2}}}{3a}
\end{align*}
$\quad\quad\quad\quad$

\begin{align*}
p_4&amp;=\sqrt{\frac{b^2}{4a^2}-\frac{2c}{3a}+p_3}
\\\\
p_5&amp;=\frac{b^2}{2a^2}-\frac{4c}{3a}-p_3
\\\\
p_6&amp;=\frac{-\frac{b^3}{a^3}+\frac{4bc}{a^2}-\frac{8d}{a}}{4p_4}
\end{align*}

Then: $$\begin{align}
x&amp;=-\frac{b}{4a}-\frac{p_4}{2}-\frac{\sqrt{p_5-p_6}}{2}
\\\\
\mathrm{or\ }x&amp;=-\frac{b}{4a}-\frac{p_4}{2}+\frac{\sqrt{p_5-p_6}}{2}
\\\\
\mathrm{or\ }x&amp;=-\frac{b}{4a}+\frac{p_4}{2}-\frac{\sqrt{p_5+p_6}}{2}
\\\\
\mathrm{or\ }x&amp;=-\frac{b}{4a}+\frac{p_4}{2}+\frac{\sqrt{p_5+p_6}}{2}
\end{align}$$

(These came from having Mathematica explicitly solve the quartic, then seeing what common bits could be pulled from the horrifically-messy formula into parameters to make it readable/useable.)
If anyone is interested, there is a quartic equation calculator with a method and complete steps to solving the roots in the link here.