Решить неравенство

Решите задачу:

$\sqrt{x+8} \ \textgreater \ x+2\\\\1)\quad \left \{ {{x+2\ \textless \ 0} \atop {x+8 \geq 0}} \right. \; \left \{ {{x\ \textgreater \ -2} \atop {x \geq -8}} \right. \; \; \to \; \; x\ \textgreater \ -2\\\\2)\quad \left \{ {{x+2 \geq 0} \atop {x+8\ \textgreater \ (x+2)^2}} \right. \; \left \{ {{x \geq -2} \atop {x+8\ \textgreater \ x^2+2x+4}} \right. \left \{ {{x \geq -2} \atop {x^2+x-4\ \textless \ 0}} \right. \; \left \{ {{x \geq -2} \atop {x\in (\frac{-1-\sqrt{17}}{2},\frac{-1+\sqrt{17}}{2})}} \right. \\\\\frac{-1-\sqrt{17}}{2}\approx -2,56\; \; ;\; \; \frac{-1+\sqrt{17}}{2}\approx 1,56$

$x\in \left [-2,\frac{-1+\sqrt{17}}{2}\right )$