Answer
$(-\infty, 0)\cup[\frac{3}{2},3)\cup[5,\infty)$
Work Step by Step
Given $\frac{2x^2-13x+15}{x^2-3x}\geq0$, we can identify it as an inequality. Graph $f(x)=\frac{2x^2-13x+15}{x^2-3x}$ as shown in the figure and we can find the solution set as $(-\infty, 0)\cup[\frac{3}{2},3)\cup[5,\infty)$
