Решите неравенство: 8/(2x-3)^2-10/(2x-3)-3<=0.

\[\frac{- 12x^{2} + 16x + 11}{(2x - 3)^{2}} \leq 0\]

\[\frac{12x^{2} - 6x - 11}{(2x - 3)^{2}} \geq 0\]

\[\frac{12 \cdot \left( x + \frac{1}{2} \right)\left( x - 1\frac{5}{6} \right)}{4 \cdot (x - 1,5)^{2}} \geq 0\]

\[Ответ:\ \ ( - \infty; - 0,5\rbrack \cup \left\lbrack 1\frac{5}{6}; + \infty \right).\]