There most certainly is, but its ugly, complicated, and not worth memorizing. People know about it and have quoted or cited it for you, but really they would never use it. If you want something actually useful for pen-and-paper solutions, you might want to understand the actual theory behind the solution. I will provide one method below.

The Quartic Formula is just the end result of this methodology, written in terms of the original coefficients. Because of that, the method is far easier to remember than the formula, which is why I find it annoying when people cite just the formula and tell you, "dont bother, use a computer instead." A pen-paper solution is not complicated, it's just time consuming.

Understanding how its done - even if you never use it - expands your brain and your understanding, allows you to implement it in programming, and allows you to recreate it whenever you may need it instead of hyper reliance on computers to always be there for you, which, in my opinion, makes for a poor mathematician.

There are three methods for solving Quartics that I both know and know of:

- Descartes' Quadratic Factorization
- Euler's Method
- Ferrari's Method

If anyone knows of more, please let me know.

Ferrari's Method is historically the first method discovered. Euler's Method looks a lot like Cardano's Method for the Cubic and was probably modeled after the same approach. But I am partial to Descartes' Quadratic Factorization technique. Its a relatively simple process to follow and is what I will be using below. If you want to see how the others work, let me know.

All of the above methods start out the same: depression (removing the $n-1$ degree term, in this case the cubic term) and normalization (making the lead coefficient 1, i.e. making the polynomial monic).

They all end up roughly the same place too: solving a cubic equation. So you need to be prepared for that. I recommend you brush up on that; I wont explain solving the cubic here but will only refer to it.

**Depress and Normalize**

Given the arbitrary quartic:
$$ax^4 + bx^3 + cx^2 + dx + e =0$$
Where $a,b,c,d,e\in\mathbb{R}$.

We must convert this quartic into a depressed monic quartic. First we depress it by substituting $x = z-\frac{b}{4a}$. We arrive at the quartic:
$$az^4 + Bz^2 + Cz + D =0$$
For some $B,C,D\in\mathbb{R}$. This quartic, now in $z$, is merely a horizontal shift from the original quartic in $x$. All points have shifted, therefore the roots have shifted, but by the same constant. Also, $B,C,D$ are computed from the previous $a,b,c,d,e$ and have no dependence on $x,z$. Whats important is that there is no $z^3$ term. Next we divide through by the lead coefficient $a$ to make this quartic monic (normalize).
$$z^4 + pz^2 + qz + r = 0\;\;\;\;\;\;\;\;\;\;\;\;(1)$$

For some $p,q,r\in\mathbb{R}$. This has no effect on the root positions; all the lead coefficient did was scale up the non-zero values. There is absolutely no loss in generality with respect to the zeros. All of these constants, $p,q,r$, can be computed from the original coefficients, $a,b,c,d,e$. There is still no cubic term and now there is no lead coefficient either.

What might be interesting to note is what has happened to the polynomial. We started out with 5 arbitrary constants and have reduced it to 3, by normalizing the lead and removing the cubic term. We originally had arbitrary values of $a,b,c,d,e\in\mathbb{R}$, and now we have arbitrary values $p,q,r\in\mathbb{R}$. Although the latter three are computed from the original five, they have arbitrary values, and there is no loss in generality. This is a significant simplification of the problem. The non-existence of the cubic term will prove vital.

Everything so far has simply been the setup: writing the polynomial in reduced monic form. Recall all of the quartic methods accomplish at least this much. Next we implement Descartes Factorization method.

**Descartes Factorization Method**

We must assume that all of the coefficients are real, $p,q,r\in\mathbb{R}$. This is a required condition to make the methodology work. The reason is because now all solutions with non-zero imaginary components come in complex conjugate pairs. Big deal? It allows us to group two solutions together, even if they are purely real, into quadratic factors with real coefficients. We know that *all* real-valued monic quartics can be factored into:
$$(z^2 + mz+n)(z^2+sz+t)=0\;\;\;\;\;\;\;\;\;\;\;\;(2)$$
Where $m,n,s,t\in\mathbb{R}$. Clearly the task at hand is to find the values of these constants. If we can convert (1) into (2) by determining quadratic factors that satisfies the equation, we can find the roots with the quadratic formula applied to those quadratic factors.

What we do is distribute these quadratic factors out into normal polynomial form and you get:
$$z^4 + (m+s)z^3 + (t+n+ms)z^2 + (mt+ns)z + nt = 0$$
Which we compare term-by-term to the depressed monic in (1). You get this system of equations:
$$ m+s=0 \\ n+t+ms=p \\mt+ns=q \\ nt=r$$
Its plain to see then that $s = -m$ and that $t=\frac{r}{n}$ from the first and fourth equations. Our quadratic factors can be rewritten:
$$(z^2 + mz+n)(z^2-mz+\frac{r}{n})=0\;\;\;\;\;\;\;\;\;\;\;\;(3)$$
We only have two unknowns in this factorization: $m$ and $n$. The remaining two of the four simultaneous equations (the second and the third equation) can be rewritten:
$$ \frac{r}{n}+n-m^2 = p \\ m(\frac{r}{n} -n)=q$$
By moving the $m$ to the right-hand side:
$$ \frac{r}{n}+n = p+m^2 \\ \frac{r}{n} -n=\frac{q}{m}$$

From here we form two new equations by adding and subtracting the previous two. By adding we get:
$$ 2\frac{r}{n} = p+m^2 + \frac{q}{m}$$
By subtracting we get:
$$ 2n = p+m^2 -\frac{q}{m}$$
Notice on the left hand side that both of these equations can be solved for $n$ and $\frac{r}{n}$ readily in terms of $m$, both of which appear in the quadratic factors of (3). These can be utilized later once we know the $m$ to complete the quadratic factor.

We can find the $m$ by taking these latest two equations and multiplying them, therefore eliminating the unknown $n$. Notice that $n$ is in the numerator in one and in the denominator in the other. Thus:
$$ 4r = (p+m^2)^2 - (\frac{q}{m})^2 = m^4 + 2pm^2 + p^2 - \frac{q^2}{m^2}$$

Thus we are essentially down to $m$ as our last unknown. Everything else is known in terms of $m$, and $m$ is the only unknown in the above equation. One unknown, one remaining equation. Multiply the equation through by $m^2$ and rearrange:
$$ m^6 + 2pm^4 + (p^2-4r)m^2 - q^2 = 0$$
This is still rather ugly. It's a sextic, not a quartic, which is worse. But notice that the powers of $m$ are all even. We can substitute $w=m^2$ and arrive at:
$$ w^3 + 2pw^2 + (p^2 - 4r)w - q^2 =0$$

And thus we are essentially done. We are left with a **cubic** polynomial in $w$, which is solvable with its own techniques. Techniques which I only assume you already know about if you are trying to solve quartics. Just like with quartics, as you know already, there do exist cubic formulae, but I do recommend learning the methods behind them.

If you need help with cubics, I recommend Cardano's method (the original solution) or Vieta's Trigonometric Solution (my preferred). There is also Completing the Cube, a nice proof of concept but Id never use it. Feel free to ask a separate question for a cubic and I will be happy to answer.

The point is that the problem has been reduced from that of finding the roots of a quartic to that of finding the roots of a cubic. A simpler problem! That's usually how it goes. All quartic root finding methods require finding the roots of a cubic first, obvious or not. Just as finding the roots of a cubic involve solving a quadratic. Hope this works for you.

Anyway, solve the cubic in $w$. Then with that recall the quadratic substitution $w=m^2$ and solve for $m$. Then recall that we previously had equations for $ 2n$ and $2\frac{r}{n}$. And now you know all of the unknown terms in the quadratic factorization $(z^2 + mz + n)(z^2 - mz + \frac{r}{n})=0$.

Still not done. Each of these quadratic factors must now be solved using the quadratic formula, and you have solutions in $z$. This solves the depressed monic quartic we started Descartes Quadratic Factorization method with.

**Finally**

Dont forget about the original quartic we had at the very beginning, prior to the depression and normalization. We had introduced a horizontal shift of $x = z-\frac{b}{4a}$. Doing this last bit will solve the original quartic in terms of $x$, which is the solution you want.

You are going to arrive at a set of solutions when done. Be sure to check your answers. You may have redundant or superfluous solutions. Some redundant solutions may be written in very different algebraic ways, but will represent the same numeric value.

If you express the final answer of $x$ in terms of the original $a,b,c,d,e$ you will just have the same "quartic formulae" that other people are citing you. The expression will of course be slightly different depending on which of the quartic methods you employ.

**Concerns**

If youre concerned about the assumption that coefficients $p,q,r$ are real, dont be. All it means is that $a,b,c,d,e$ are real, which is usually a good assumption. In fact, we can generalize. The values $p,q,r$ can be made complex, implying only that the original quartic has complex $a,b,c,d,e$. It also means you'll have to solve a cubic with complex coefficients. This is doable and the math still works fine.