Решите неравенство 2х^2-8х_>0 Решите систему Х+4_>0 Х^2-х-56<0

Решите задачу:

$1)\; \; 2x^2-8x\ \textgreater \ 0\\\\2x(x-4)\ \textgreater \ 0\; \; \; \; \; +++(0)---(4)+++\\\\x\in (-\infty ,0)\cup (4,+\infty )$

$2)\; \; \left \{ {{x+4\ \textgreater \ 0} \atop {x^2-x-56\ \textless \ 0}} \right. \; \left \{ {{x\ \textgreater \ -4} \atop {(x+7)(x-8)\ \textless \ 0}} \right. \; \left \{ {{x\in (-4,+\infty )} \atop {x\in (-7,8)}} \right. \; \Rightarrow \; \; x\in (-4,8)\\\\x^2-x-56=0\; \to x_1=-7,\; x_2=8\; \; (teor.\; Vieta)\\\\+++(-7)---(8)+++\\$