Решите неравенство: (x^2+2x)^2-11*(x^2+2x)+24<=0.

\[t = x^{2} + 2x\]

\[t^{2} - 11t + 24 \leq 0\]

\[(t - 8)(t - 3) \leq 0\]

\[3 \leq t \leq 8 \Longrightarrow 3 \leq x^{2} + 2x \leq 8\]

\[\left\{ \begin{matrix} x^{2} + 2x \geq 3 \\ x^{2} + 2x \leq 8 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} x^{2} + 2x - 3 \geq 0 \\ x^{2} + 2x - 8 \leq 0 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} (x - 1)(x + 3) \geq 0 \\ (x - 2)(x + 4) \leq 0 \\ \end{matrix} \right.\ \]

\[Ответ:\lbrack - 4; - 3\rbrack \cup \lbrack 1;2\rbrack\text{.\ }\]