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.)
Great question I believe! You probably know this but this is said in MacTutor's Quadratic, cubic and quartic equation.
"After Tartaglia had shown Cardan how to solve cubics, Cardan
encouraged his own student, Lodovico Ferrari, to examine quartic
equations. Ferrari managed to solve the quartic with perhaps the most
elegant of all the methods that were found to solve this type of
problem. Cardan published all 20 cases of quartic equations in Ars
Magna. Here, again in modern notation, is Ferrari's solution of the
case: x4 + px2 + qx + r = 0. First complete the square to obtain
[ ... ]
In the years after Cardan's Ars Magna many mathematicians contributed
to the solution of cubic and quartic equations. Viète, Harriot,
Tschirnhaus, Euler, Bezout and Descartes all devised methods.
Tschirnhaus's methods were extended by the Swedish mathematician E S
Bring near the end of the 18th Century."
I just wanted to put it here to emphasize how "non-trivial" the problem of solving quartic equations can be. (I believe that you are aware of methods of finding rational roots, Uspensky's Theory of Equations is a great reference for that; but the general problem is definitely non trivial).
Here are many of this methods explained. I also found some of them here.
Best Answer
We can assume that $a=1$ by dividing by it, so start with $$ x^4 + bx^3 + cx^2 + dx + e = 0 $$ (there's no need for $f$, either).
We will use a more general substitution known as a Tschirnhaus substitution: put $$ y = x^2 + \lambda x + \mu . $$ Then $$ y^2 = x^4 + 2\lambda x^3 + 2(\mu+\lambda^2) x^2 + 2\lambda\mu x + \mu^2 = (2\lambda-b)x^3 + (\lambda^2+2\mu-c)x^2 + (2\lambda\mu-d)x + (\mu^2-e) \\ y^3 = \dotsb = (-b^3 + 2 b c - d + 3 b^2 \lambda - 3 c \lambda - 3 b \lambda^2 + \lambda^3 - 3 b \mu + 6 \lambda \mu) x^3 + (-b^2 c + c^2 + b d - e + 3 b c \lambda - 3 d \lambda - 3 c \lambda^2 - 3 c \mu + 3 \lambda^2 \mu + 3 \mu^2) + \dotsb \\ y^4 = \dotsb , $$ and then since $1,y,y^2,y^3,y^4$ give five equations with four coefficients, we can find a linear combination so that they add up to zero, leaving us with the equation $$ y^4 + (-b^2 + 2 c + b \lambda - 4 \mu) y^3 + (c^2 - 2 b d + 2 e - b c \lambda + 3 d \lambda + c \lambda^2 + 3 b^2 \mu - 6 c \mu - 3 b \lambda \mu + 6 \mu^2) y^2 + ()y + () = 0 , $$ where we're not interested in the last two coefficients. (This effectively calculates the resultant of the two polynomials $x^4+bx^3+cx^2+dx+e$ and $x^2+\lambda x-\mu-y$, which is generally best done with a computer.)
To eliminate the first two coefficients, we need $$ -b^2 + 2 c + b \lambda - 4 \mu = 0 \\ c^2 - 2 b d + 2 e - b c \lambda + 3 d \lambda + c \lambda^2 + 3 b^2 \mu - 6 c \mu - 3 b \lambda \mu + 6 \mu^2 = 0 . $$ The first of these is linear, the second quadratic, so it is easy to solve both simultaneously: we find $$ \lambda = \frac{3 b^3 - 10 b c + 12 d \pm 2 \Delta}{3b^2-8c}, \qquad \mu = \frac{2 b^2 c - 8 c^2 + 6 b d \pm b \Delta}{6b^2-16c} , $$ where $\Delta^2 = 6 b^3 d - 28 b c d - 2 b^2 (c^2 - 6 e) + 4 (2 c^3 + 9 d^2 - 8 c e) $. I've left out a few of the details of the calculation, but they should be easy, if tedious, to fill in.