Answer
$x_{1}=0$ and $x_{2}=-16$
Work Step by Step
$(2x-3)(x+6)=(x-9)(x+2)$
$2x(x+6)-3(x+6)=x(x+2)-9(x+2)$
$2x^2+12x-3x-18=x^2+2x-9x-18$
$2x^2+9x-18=x^2-7x-18$
$2x^2-x^2+9x+7x-18+18=0$
$x^2+16x=0$
$x(x+16)=0$
Therefore the solutions are $x_{1}=0$ and $x_{2}=-16$
