Answer
$(-3,0)\cup(2,\frac{9}{2}]$
Work Step by Step
$f(x)=\frac{x^2+3x+2x^2-4x-3x^2-3x+18}{x(x-2)(x+3)}=\frac{-4x+18}{x(x-2)(x+3)}$
$f(x)=\frac{-2(2x-9)}{x(x-2)(x+3)}\geq0$
Use cut points $-3,0,2,4,5$ to make a table as shown.
Drawn conclusion from the table as $(-3,0)\cup(2,\frac{9}{2}]$
