Решите неравенство: х2>2х+15

$x^2\ \textgreater \ 2x+15\\ x^2-2x\ \textgreater \ 15\\ x^2-2x+1-1\ \textgreater \ 15\\ (x-1)^2\ \textgreater \ 16\\ |x-1|\ \textgreater \ 4\\ \left[\begin{array}{ccc}x-1\ \textgreater \ 4\\ x-1\ \textless \ -4\end{array}\right\Rightarrow \left[\begin{array}{ccc}x\ \textgreater \ 5\\ x\ \textless \ -3\end{array}\right$

Ответ: $x \in (-\infty;-3)\cup(5;+\infty)$