Решим неравенство:
$$\frac{-13}{(x-12)^2 - 5} \ge 0$$
Т.к. $$-13 < 0$$, то $$(x-12)^2 - 5 < 0$$.
$$(x-12)^2 < 5$$
$$-\sqrt{5} < x-12 < \sqrt{5}$$
$$12-\sqrt{5} < x < 12+\sqrt{5}$$
$$12-2.236 < x < 12+2.236$$
$$9.764 < x < 14.236$$
Ответ: $$12-\sqrt{5} < x < 12+\sqrt{5}$$