$\left|x^{2}-x-6\right|=x+2$, then
Case I : $x^{2}-x-6 < 0$
$\Rightarrow(x-3)(x+2) < 0$
$\Rightarrow-2 < x < 3$
In this case, the equation becomes $x^{2}-x-6=-x-2$
or $x^{2}-4=0$
$\therefore \mathrm{x}=\pm 2$
Clearly, $x=2$ satisfies the domain of the equation in this case. So, $x=2$ is a solution.
Case II : $x^{2}-x-6 \geq 0$
So, $x \leq-2$ or $x \geq 3$
In this case, the equation becomes
$x^{2}-x-6=0=x+2$
i.e., $x^{2}-2 x-8=0$ or $x=-2,4$
Both these values lie in the domain of the equation in this case, so $x=-2,4$ are the roots.
Hence, roots are $x=-2,2,4$.