Помогите решить уравнение. Заранее СПАСИБО!!!

-4} \atop {\sqrt{x^2-16}+1<4}} \right.\\\\ \left \{ {{\sqrt{x^2-16}>-5} \atop {\sqrt{x^2-16}<3}} \right.\\\\ \left \{ {{x^2-16>25}} \atop {x^2-16<9}} \right. \\\\ \left \{ {{x^2>31} \atop {x^2<25}} \right. " alt="1)x^2-16 \geq 0\\ (-\infty;-4) \ \cup \ (4;\infty)\\\\ 2)x^2+2\sqrt{x^2-16}-31<0\\ (\sqrt{x^2-16}+1)^2-16<0\\ (\sqrt{x^2-16}+1)^2<16\\\\ \left \{ {{\sqrt{x^2-16}+1>-4} \atop {\sqrt{x^2-16}+1<4}} \right.\\\\ \left \{ {{\sqrt{x^2-16}>-5} \atop {\sqrt{x^2-16}<3}} \right.\\\\ \left \{ {{x^2-16>25}} \atop {x^2-16<9}} \right. \\\\ \left \{ {{x^2>31} \atop {x^2<25}} \right. " align="absmiddle" class="latex-formula">
Откуда получаем решения
$(-5;-4] \ \cup \ [4;5)$