Solution. Let $a=\log _{2} 3$. Then the condition of the problem will turn into the inequality

$$
x^{2}+2\left(a-\frac{2}{a}+1\right) x-\left(2 a+\frac{12}{a}+11\right) \leqslant 0
$$

Considering that $a \in\left(\frac{3}{2}, 2\right)$, we get $x \in\left[-2 a-3, \frac{4}{a}+1\right]$. Since $-7<-2 a-3<-6,3<\frac{4}{a}+1<4$, the integer solutions will be the numbers $-6,-5,-4,-3,-2,-1,0,1,2,3$.

Answer: $-15 .(\mathrm{B})$

$$
\begin{array}{|l|l|l|l|l|l|l|l|}
\hline \mathbf{A}-20 & \mathbf{B}-15 & \mathbf{C} & -10 & \mathbf{D} & -5 & \mathbf{E} 0 & \mathbf{F} \\
\hline
\end{array}
$$
5.3. Find the sum of all integer values of the argument $x$ for which the corresponding values of the function

$$
y=x^{2}+x\left(\log _{2} 36-\log _{3} 16\right)-\log _{2} 9-4 \log _{3} 8
$$

do not exceed 11.