Решите неравенство: Варианты ответов: A) (2;6) B) (4;6] C) (4;5) D) (2;3)

Решите задачу:

$log_{\frac{1}{2}}(x+4)+log_2(2x^2-11x+12) \leq log_2(x+3)\\\\ODZ:\; \; \left \{ {{x+4\ \textgreater \ 0,\; x+3\ \textgreater \ 0} \atop {2x^2-11x+12\ \textgreater \ 0}} \right. \; \left \{ {{x\ \textgreater \ -3} \atop {2(x-\frac{3}{2})(x-4)\ \textgreater \ 0}} \right. \; \; \to \\\\x\in (-3,\frac{3}{2})\cup (4,+\infty )\\\\-log_2(x+4)+log_2(2x^2-11x+12) \leq log_2(x+3)\\\\log_2(2x^2-11x+12) \leq log_2(x+3)+log_2(x+4)\\\\2x^2-11x+12\leq (x+3)(x+4)\\\\2x^2-11x+12 \leq x^2+7x+12\\\\x^2-18x \leq 0\\\\x(x-18) \leq 0\\\\+++[\, 0\, ]---[\, 18\, ]+++\\\\x\in [\, 0,18\, ]$

$\left \{ {{x\in[\, 0,18\, ]} \atop {x\in (-3,\frac{3}{2})\cup (4,+\infty )}} \right. \; \; \; \; \; x\in (4,18\, ]$