Solve quartic equation $a^4-6a^2b-8ac-3b^2=0$ Please help me to find roots of this quartic equation for a:
$$a^4-6a^2b-8ac-3b^2=0$$
Wolfram Alpha gave this result.
But may be there is simple way to get all a?

Rewrite the equation by completing the squares
\begin{align}
a^4-6a^2b-8ac-3b^2
=(a^2+s)^2 -( 2s+6b)\left( a +\frac{4c}{2s+6b} \right)^2\tag1\\
\end{align}
where $s$ happens to satisfy $(s+b)^3=8(c^2-b^3)$, or
$s=2\sqrt[3]{c^2-b^3}-b$.
For notational convenience, denote
$$p=\frac12\sqrt{2s+6b}= \sqrt{\sqrt[3]{c^2-b^3}+b }\tag2$$
and then $s=2p^2-3b$.
Substitute them into (1) and then factorize to get
$$\left(a^2+2pa+2p^2-3b-\frac{2c}p\right) \left(a^2-2pa+2p^2-3b+\frac{2c}p\right)=0
$$
which yields the solutions
$$a=p\pm \sqrt{ 3b-p^2+\frac{2c}p },\>\>\> -p\pm \sqrt{ 3b-p^2-\frac{2c}p } $$
with $p$ given by (2).
