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 $ax^4 + bx^3+c x^2 + d x +e=0$ 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&=2c^3-9bcd+27ad^2+27b^2e-72ace
\\\\
p_2&=p_1+\sqrt{-4(c^2-3bd+12ae)^3+p_1^2}
\\\\
p_3&=\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&=\sqrt{\frac{b^2}{4a^2}-\frac{2c}{3a}+p_3}
\\\\
p_5&=\frac{b^2}{2a^2}-\frac{4c}{3a}-p_3
\\\\
p_6&=\frac{-\frac{b^3}{a^3}+\frac{4bc}{a^2}-\frac{8d}{a}}{4p_4}
\end{align*}
Then: $$\begin{align}
x&=-\frac{b}{4a}-\frac{p_4}{2}-\frac{\sqrt{p_5-p_6}}{2}
\\\\
\mathrm{or\ }x&=-\frac{b}{4a}-\frac{p_4}{2}+\frac{\sqrt{p_5-p_6}}{2}
\\\\
\mathrm{or\ }x&=-\frac{b}{4a}+\frac{p_4}{2}-\frac{\sqrt{p_5+p_6}}{2}
\\\\
\mathrm{or\ }x&=-\frac{b}{4a}+\frac{p_4}{2}+\frac{\sqrt{p_5+p_6}}{2}
\end{align}$$
(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.)
Best Answer
The main problem you're going to have is a general forth degree polynomial will not be one-to-one, so you'll have to be careful to restrict the domain of your function. Take for example the most basic quartic:
$$ f(x)=x^4 $$ If $f(x)=1$, what is $x$? Well (if you're only considering real numbers), $x$ could be either $1$ or $-1$. So which is it? If you restrict your domain to only positive numbers, then the inverse is $1$.
The problem becomes more complex, of course, when you consider the "general" case, since the function will look like a "W" in general (you might have 4 possible solutions!)
If you're okay with inexact answers (as this is for programming, I'm guessing this might be the case?), you can always solve the equation $y=f(x)$ for $x$ by using a numerical method such as Newton's method. Of course, yet again, you'll have to worry about multiple solutions.