Answer
$[-\frac{24}{7},-2)\cup(0,3)$
Work Step by Step
$r(x)=\frac{x^2-3x+3x^2+6x-4x^2+4x+24}{x(x+2)(x-3)}=\frac{7x+24}{x(x+2)(x-3)}\leq0$
Use cut points $-24/7(-3.4),-2,0,3$ to make a table as shown.
Draw conclusion from the table $[-\frac{24}{7},-2)\cup(0,3)$
