Is there formula that solves quartic equation $ax^4+bx+c=0$ In general form, a quartic equation is $ax^4+bx^3+cx^2+dx+e=0$. I was thinking of the quartic equation of the form
$$ax^4+bx+c=0$$
which resembles a depressed cubic equation.
Does anyone know a formula that solves this type of depressed quartic equation,  simlilar to the formula
$$\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
for the quadratic equation $ax^2+bx+c=0$, or
$$\sqrt[3]{\frac{-q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}+\sqrt[3]{\frac{-q}{2}-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}$$ for the cubic equation $x^3+px +q=0$.

The quartic equation of the form
$$ax^4+bx+c=0$$
has the following formulas for its four roots
$$x= -s \pm \sqrt{ \frac b{4as}-s^2},\>\>\> s \pm \sqrt{ -\frac b{4as}-s^2} $$
where $s^2$ satisfies the cubic equation $s^6 -\frac c{4a} s^2-\left(\frac b{8a}\right)^2=0$. Take the example of
$$x^4+2x+\frac12=0$$
for which $s^6 -\frac 18 s^2-\frac1{16}=0$ or  $s^2=\frac1{2}$.
Plug $s$ into the formulas above to obtain
$$x=\frac{-1\pm\sqrt{\sqrt2-1}}{\sqrt2},\>\>\> 
\frac{1\pm i\sqrt{\sqrt2+1}}{\sqrt2} $$
