Решите неравенство (х+2)(х-2)-2(х-1)≥23-2х

$\textup{(x+2)(x-2)-2(x-1)} \geq \textup{23-2x} \\ \textup{x}^2-4-2\textup{x}+2 \geq 23-2\textup{x} \\ \textup{x}^2-2\textup{x}-2 \geq 23-2\textup{x} \\ \textup{x}^2-2\textup{x}+2\textup{x}-2-23 \geq 0 \\ \textup{x}^2-25 \geq 0 \\(\textup{x}+5)(\textup{x}-5) \geq 0$

_______+______ $-5$ ______-______ $5$ ______+_______

Ответ: $\textup{A}) ~\textup{x}\in(-\infty;-5]\textup{ U }[5;+\infty)$