Answer: $-13-\sqrt{57} \approx-20.55$.

Solution. Let $f(x)=-\log _{2}(x+120)$. Then, taking into account the condition $71-2 x \sqrt{32-2 x}>0$, the original inequality can be rewritten as

$$\frac{2 f(-2 x \sqrt{32-2 x})+\left|3 f\left(x^{2}-2 x-112\right)-f(-2 x \sqrt{32-2 x})\right|}{\left(2 \log _{2} 7\right) \cdot f(-2 x \sqrt{32-2 x})-5 f(-2 x \sqrt{32-2 x}-49)} \geqslant 0$$

For the function $h(x)=-2 x \sqrt{32-2 x}$, we have

$$h^{\prime}(x)=-2 \sqrt{32-2 x}+\frac{2 x}{\sqrt{32-2 x}}=\frac{6 x-64}{\sqrt{32-2 x}}$$

the point of global minimum of the function $h(x)$ will be $x_{*}=\frac{32}{3}, h\left(x_{*}\right)=-\frac{64}{3} \sqrt{\frac{32}{3}} \approx-69.67$. For the function $g(x)=x^{2}-2 x-112$, it is obviously true that $g(x) \geqslant-113$. Thus, for all admissible $x$, the expressions

$$-2 x \sqrt{32-2 x}, x^{2}-2 x-112,-2 x \sqrt{32-2 x}-49$$

lie in the set $(-119 ;+\infty)$, on which the function $f(x)$ is decreasing and negative. Then

$\left(2 \log _{2} 7\right) \cdot f(-2 x \sqrt{32-2 x})-5 f(-2 x \sqrt{32-2 x}-49)<\left(2 \log _{2} 7-5\right) \cdot f(-2 x \sqrt{32-2 x}-49)<0$ for all admissible $x$, so the original inequality is equivalent to the inequality

$$\left|3 f\left(x^{2}-2 x-112\right)-f(-2 x \sqrt{32-2 x})\right| \leqslant-2 f(-2 x \sqrt{32-2 x})$$

which, in turn, is equivalent to the system

$$\left\{\begin{array}{l} 3 f\left(x^{2}-2 x-112\right) \leqslant-f(-2 x \sqrt{32-2 x}) \\ f\left(x^{2}-2 x-112\right) \geqslant f(-2 x \sqrt{32-2 x}) \end{array}\right.$$

The first inequality of this system is satisfied for all admissible $x$ due to the negativity of $f(x)$, the second one, due to its monotonicity, is equivalent to the inequality $x^{2}-2 x-112 \leqslant-2 x \sqrt{32-2 x}$. It can be rewritten as $(x+\sqrt{32-2 x})^{2} \leqslant 144$, which is equivalent to the system

$$\left\{\begin{array}{l} \sqrt{32-2 x} \leqslant 12-x \\ \sqrt{32-2 x} \geqslant-12-x \end{array}\right.$$

the solutions of which will be $x \in[-13-\sqrt{57} ; 8]$.