#### Answer

$x_{1}=0$ and $x_{2}=-15$

#### Work Step by Step

$(2x-3)(x+8)=(x-6)(x+4)$
$2x(x+8)-3(x+8)=x(x+4)-6(x+4)$
$2x^2+16x-3x-24=x^2+4x-6x-24$
$2x^2+13x-24=x^2-2x-24$
$2x^2-x^2+13x+2x-24+24=0$
$x^2+15x=0$
$x(x+15)=0$
Therefore the solutions are $x_{1}=0$ and $x_{2}=-15$