If $\left|x^{2}-x-6\right|=x+2$, then the values of $x$ are

$\left|x^{2}-x-6\right|=x+2$, 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 $x=-2,2,4$.