Is there a general formula for solving Quartic (Degree $4$) equations? There is a general formula for solving quadratic equations, namely the Quadratic Formula, or the Sridharacharya Formula:

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

For cubic 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 [Quartic Equations]?
$$ ax^4 + bx^3 + cx^2 + dx + e = 0 $$
How about for higher degrees? If not, why not?

Contrary to common opinion, the resolution of the quartic with real coeficients is not so difficult.
I take for granted that depletion and reduction of the equation by a linear change of variable
$$\alpha x^4+\beta x^3+\gamma x^2+\delta x+\epsilon=0\to x^4+px^2+qx+r=0$$ are known.
Then we factor this depleted quadrinomial as
$$(x^2+ax+b)(x^2-ax+c)=x^4+(-a^2+b+c)x^2+a(c-b)x+bc$$ and identify the coefficients,
$$-a^2+b+c=p,\\a(c-b)=q,\\bc=r.$$
We can express $c\pm b$ in terms of $a$, solve for $b,c$ then form the product $4a^2bc$:
$$c-b=\frac qa,\\c+b=p+a^2,$$ hence
$$4a^2bc=\left(a(p+a^2)-q\right)\left(a(p+a^2)+q\right)=a^2(p^2+a^2)^2-q^2=4a^2r.$$
This is a cubic equation in $a^2$, and fortunately it always has a positive solution.
Then after resolution, from $a$ we obtain $b$ and $c$, and the roots of the two quadratic trinomials.
