$\left|x^{2}-x-6\right|=x+2$, 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 $x^{2}-4 =0$
$\therefore x=\pm 2$
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 $x \geq 3$
Then, equation reduces to $x^{2}-x-6=0=x+2$
i.e., $x^{2}-2 x-8=0$ or $x=-2,4$
Both these values lies in the domain of the equation in this case,
so $x=-2,4$ are the roots.
Hence, roots are $x=-2,2,4$.