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. $x \in\left(-\infty ; \frac{1-\sqrt{17}}{2}\right] \cup[0 ;+\infty)$.

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

$$
\begin{aligned}
& {\left[\begin{array} { l } 
{ x ^ { 3 } - 2 x ^ { 2 } + 2 \geqslant 2 - 3 x , } \\
{ x ^ { 3 } - 2 x ^ { 2 } + 2 \leqslant - 2 + 3 x }
\end{array} \Leftrightarrow \left[\begin{array} { l } 
{ x ^ { 3 } - 2 x ^ { 2 } + 3 x \geqslant 0 , } \\
{ x ^ { 3 } - 2 x ^ { 2 } - 3 x + 4 \leqslant 0 }
\end{array} \Leftrightarrow \left[\begin{array}{l}
x\left(x^{2}-2 x+3\right) \geqslant 0, \\
(x-1)\left(x^{2}-x-4\right) \leqslant 0
\end{array} \Leftrightarrow\right.\right.\right.} \\
& \Leftrightarrow\left[\begin{array}{l}
x \geqslant 0 \\
x \in\left(-\infty ; \frac{1-\sqrt{17}}{2}\right] \cup\left[1 ; \frac{1+\sqrt{17}}{2}\right] \Leftrightarrow x \in\left(-\infty ; \frac{1-\sqrt{17}}{2}\right] \cup[0 ;+\infty) .
\end{array}\right.
\end{aligned}
$$