\(\left| {2{x^2}-9x + 15} \right| \ge 20.\)
Неравенство вида \(\left| {f\left( x \right)} \right| \ge a\) при \(a > 0\) равносильно совокупности: \(\left[ {\begin{array}{*{20}{c}}{f\left( x \right) \ge a,\,\,}\\{f\left( x \right) \le -a.}\end{array}} \right.\)
\(\left| {2{x^2}-9x + 15} \right| \ge 20\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{2{x^2}-9x + 15 \ge 20,\,}\\{2{x^2}-9x + 15 \le -20}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{2{x^2}-9x-5 \ge 0,\,}\\{2{x^2}-9x + 35 \le 0}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x \in \left( {-\infty ;\;-\dfrac{1}{2}} \right] \cup \left[ {5;\; + \infty } \right).\)
Ответ: \(\left( {-\infty ;\;-\dfrac{1}{2}} \right] \cup \left[ {5;\; + \infty } \right).\)