Answer
$x_{1}=0$ and $x_{2}=\dfrac{1}{2}$
Work Step by Step
$(x-3)(3x+4)=(x+2)(x-6)$
$x(3x+4)-3(3x+4)=x(x-6)+2(x-6)$
$3x^2+4x-9x-12=x^2-6x+2x-12$
$3x^2-5x-12=x^2-4x-12$
$3x^2-x^2-5x+4x-12+12=0$
$2x^2-x=0$
$x(2x-1)=0$
Therefore the solutions are $x_{1}=0$ and $x_{2}=\dfrac{1}{2}$
