Корен из x^2+2x-8> x-4 решите неравенство плеазз

Решите задачу:

$\sqrt{x^2+2x-8} \ \textgreater \ x-4\; \; \Leftrightarrow \\\\ \left \{ {{x-4\ \textless \ 0} \atop {x^2+2x-8 \geq 0}} \right. \; \; ili\; \; \left \{ {{x-4 \geq 0} \atop {x^2+2x-8\ \textgreater \ (x-4)^2}} \right. \\\\ \left \{ {{x\ \textless \ 4} \atop {(x+4)(x-2) \geq 0}} \right. \; \; ili\; \; \left \{ {{x \geq 4} \atop {x^2+2x-8\ \textgreater \ x^2-8x+16}} \right. \\\\ \left \{ {{x\ \textless \ 4} \atop {x\in (-\infty ,-4]\cup [2,+\infty )}} \right. \; \; ili\; \; \left \{ {{x \geq 4} \atop {x\ \textgreater \ 2,4}} \right. \\\\x\in [\, 2,4)\; \; \; ili\; \; \; x \geq 4\; \; (\; x\in [\, 4,+\infty )\; )\\\\Otvet:\; \; x\in [\, 2,+\infty )\; .$