A radical inequality

Domain. $x\ge-2$.

Case $x<0$. $\sqrt{x+2}\ge0>x$, so no solution.

Case $x\ge0$. Both sides $\ge0$, so we can square: $x+2\le x^2$, i.e. $x^2-x-2\ge0$, i.e. $(x-2)(x+1)\ge0$. On $x\ge0$, this gives $x\ge2$.

Conclusion. $x\ge2$.