We can reduce the problem of solving the general quartic to merely solving a quadratic. Given,
$$x^4+ax^3+bx^2+cx+d=0$$
then,
$$x_{1,2} = -\tfrac{1}{4}a+\tfrac{1}{2}\sqrt{u}\pm\tfrac{1}{4}\sqrt{3a^2-8b-4u+\frac{-a^3+4ab-8c}{\sqrt{u}}}\tag1$$
$$x_{3,4} = -\tfrac{1}{4}a-\tfrac{1}{2}\sqrt{u}\pm\tfrac{1}{4}\sqrt{3a^2-8b-4u-\frac{-a^3+4ab-8c}{\sqrt{u}}}\tag2$$
where,
$$u = \frac{a^2}{4} +\frac{-2b+v_1^{1/3}+v_2^{1/3}}{3}$$
and the $v_i$ are the two roots of the quadratic,
$$v^2 + (-2 b^3 + 9 a b c - 27 c^2 - 27 a^2 d + 72 b d)v + (b^2 - 3 a c + 12 d)^3 = 0$$
Note: If you prefer only one cube root extraction, you can use the RHS of the relation,
$$v_1^{1/3}+v_2^{1/3} = \color{brown}{v_1^{1/3}}+\frac{b^2 - 3 a c + 12 d}{\color{brown}{v_1^{1/3}}}$$
P.S. This is essentially the method used by Mathematica, though much simplified for aesthetics.