9.032. $\left|2 x^{2}-9 x+15\right| \geq 20$.

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9.032. $\left|2 x^{2}-9 x+15\right| \geq 20$.
## Solution.

We have $2 x^{2}-9 x+15>0$ for $x \in R$, therefore, the original inequality is equivalent to the inequality $2 x^{2}-9 x+15 \geq 20, 2 x^{2}-9 x-5 \geq 0$, $2(x-5)\left(x+\frac{1}{2}\right) \geq 0$. From this, $x \leq-\frac{1}{2}$ or $x \geq 5$.

Answer: $\quad x \in\left(-\infty ;-\frac{1}{2}\right] \cup[5 ; \infty)$.