2. Solve the inequality $\left|x^{3}+2 x^{2}-2\right| \geqslant-2-3 x$.

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The provided text has been translated into English while preserving the original formatting and line breaks.
Answer. $x \in\left(-\infty ; \frac{-1-\sqrt{17}}{2}\right] \cup[-1 ;+\infty)$.

Solution. The given inequality is equivalent to the following system:

$$
\begin{aligned}
& {\left[\begin{array} { l } 
{ x ^ { 3 } + 2 x ^ { 2 } - 2 \geqslant - 2 - 3 x , } \\
{ x ^ { 3 } + 2 x ^ { 2 } - 2 \leqslant 2 + 3 x }
\end{array} \Leftrightarrow \left[\begin{array} { l } 
{ x ^ { 3 } + 2 x ^ { 2 } + 3 x \geqslant 0 , } \\
{ x ^ { 3 } + 2 x ^ { 2 } - 3 x - 4 \leqslant 0 }
\end{array} \Leftrightarrow \left[\begin{array}{l}
x\left(x^{2}+2 x+3\right) \geqslant 0 \\
(x+1)\left(x^{2}+x-4\right) \leqslant 0
\end{array} \Leftrightarrow\right.\right.\right.}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_df1bfc72081ab0c9a42bg-3.jpg?height=191&width=1505&top_left_y=1549&top_left_x=435)