Решите систему неравенств: x^2+x-12<=0; 8+2x<=0.

\[\left\{ \begin{matrix} x^{2} + x - 12 \leq 0\ \ \ \ \ \ \ \ \ \ \ \\ 8 + 2x \leq 0,\ \ \ \ \ \ x \leq - 4 \\ \end{matrix}\ \right.\ \]

\[x^{2} + x - 12 = 0\]

\[x_{1} + x_{2} = - 1\ \ \ \ \ \ x_{1} = - 4\]

\[x_{1}x_{2} = - 12\ \ \ \ \ \ \ \ \ x_{2} = 3\]

\[Ответ:x \in \left\{ - 4 \right\}.\]