If $\ddot{a}\le 0$ the roots of $x^2-2a|x-a|-3a^2=0$ is

$\text{ Let }a=-p\text{, then }p\ge 0\text{. Then equation becomes }$
$x^2+2p|x+p|-3p^2=0$

Case l: $x<-p.$
then $x^2-2px-5p^2=0$
$\begin{array}{r}x=\frac{\left(2p\pm \sqrt{24p^2}\right)}{2}\Rightarrow x=p\pm \sqrt{6}p\\ x=p(1+\sqrt{6})\text{ or }p(1-\sqrt{6})\end{array}$
For x, < - p, only solution is $p(1-\sqrt{6})$
$\text{ i.e., }(\sqrt{6}-1)a\text{. }$
Case II :$\mathit{x}\ge -\mathit{p}$
Then $x^2+2px-p^2=0$
$\begin{array}{l}\Rightarrow x=\frac{\left(-2p\pm \sqrt{8p^2}\right)}{2}=-p\pm \sqrt{2}p\\ \Rightarrow x=(-1+\sqrt{2})p\text{ or }(-1-\sqrt{2})a\end{array}$
As $x\ge -p,x=(-1+\sqrt{2})p=(1-\sqrt{2})a$
Hence solutions are $(\sqrt{6}-1)a\text{ and }(1-\sqrt{2})a$