The idea is basically:
Any monic polynomial can be factored as $f(x) = \prod (x - a_i)$, where $a_{1,\dots,n}$ are the roots of the polynomial.
Now if we expand such a product:
$(x - a_1)(x - a_2) = x^2 - (a_1 + a_2)x + a_1a_2$
$(x - a_1)(x - a_2)(x - a_3) = x^3 - (a_1 + a_2 + a_3)x^2 + (a_1a_2 + a_1a_3 + a_2a_3)x - a_1a_2a_3$
And so on. The pattern should be clear.
This means that finding the roots of a polynomial is in fact equivalent to solving systems like the following:
For a quadratic polynomial $x^2 - px + q$, find $a_1,a_2$, such that
$p = a_1 + a_2$
$q = a_1a_2$
For a cubic polynomial $x^3 - px^2 + qx - r$, find $a_1,a_2,a_3$, such that
$p = a_1 + a_2 + a_3$
$q = a_1a_2 + a_1a_3 + a_2a_3$
$r = a_1 a_2 a_3$
And similarly for higher degree polynomials.
Not surprisingly, the amount of "unfolding" that needs to be done to solve the quadratic system is much less than the amount of "unfolding" needed for the cubic system.
The reason why polynomials of degree 5 or higher are not solvable by radicals, can be thought of as: The structure (symmetries) of the system for such a polynomial just doesn't match any of the structures that can be obtained by combining the structures of the elementary operations (adding subtracting, multiplication, division, and taking roots).
Yes, we do have a cubic formula! By Cardan's Method...
Cardan's Method: To solve the general cubic$$x^3+ax^2+bx+c=0\tag{i}$$
Remove the $ax^2$ term by substituting $x=\dfrac {y-a}3$. Let the transformed equation be$$y^3+qy+r=0\tag{ii}$$
To solve this depressed cubic, substitute $y=u+v$ to get$$u^3+v^3+(3uv+q)(u+v)+r=0\tag{iii}$$
Put $3uv+q=0$ to get $u=-\dfrac q{3v}$ and substituting this back gives a quadratic in $v^3$. The roots of the quadratic are equal to $u^3,v^3$ respectively. And from our substitution, we get a root as$$y=\left\{-\frac r2+\sqrt{\frac {r^2}4+\frac {q^3}{27}}\right\}^{\frac 13}+\left\{-\frac r2-\sqrt{\frac {r^2}4+\frac {q^3}{27}}\right\}^{\frac 13}\tag{iv}$$
With the other two roots found with the cube roots of unity.
To find the original root of $(\text i)$, substitute $y$ into your transformation.
Best Answer
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.)