The roots of the equation  $|x^2-x-6|=x+2$

$\left|x\right|=\pm x$

$|x^2-x-6|=x+2$

Suppose that 

$\begin{array}{rcl}{x}^2-x-6& >& 0\\ & \Rightarrow & x^2-3x+2x-6>0\\ \left(x-3\right)\left(x+2\right)& >& 0\end{array}$

It means $x$ does not belong to $-2$ and $3$

$$\begin{gathered} x^2-x-6=x+2 \\ {x}^2-2x-8=0 \\ \left(x-4\right)\left(x+2\right)=0 \\ x=4,-2 \end{gathered}$$

Therefore, the roots are $-2,4$

Suppose that $x^2-x-6\le 0$, It means $x$ belong to $-2$ and $3$

$$\begin{gathered} x^2-x-6=-x-2 \\ {x}^2=4 \\ x=2,-2 \end{gathered}$$

Therefore, the roots are $-2,2,4$