Answer
$\{-1, \frac{5}{2}, \frac{3\pm i\sqrt {39}}{4} \}$
Work Step by Step
Step 1. Given $|2x^2-3x|=5$, remove the absolute sign to get $2x^2-3x=5$ and $2x^2-3x=-5$
Step 2. For $2x^2-3x=5$, we have $2x^2-3x-5=0$ or $(x+1)(2x-5)=0$ which gives $x=-1, \frac{5}{2}$
Step 3. For $2x^2-3x=-5$, we have $2x^2-3x+5=0$ which gives $x=\frac{3\pm\sqrt {1-4(2)(5)}}{2(2)}=\frac{3\pm i\sqrt {39}}{4}$
Step 4. Combine the above results, the solution set is $\{-1, \frac{5}{2}, \frac{3\pm i\sqrt {39}}{4} \}$
