Answer
$x=4$ or $x=-1$
Work Step by Step
We solve by factoring:
$x^{2}-3x-\sqrt{x^{2}-3x}=2$
$x^{2}-3x-\sqrt{x^{2}-3x}-2=0$
$[\sqrt{x^2-3x}+1][\sqrt{x^2-3x}-2]=0$
$[\sqrt{x^2-3x}+1]=0$ or $[\sqrt{x^2-3x}-2]=0$
$\sqrt{x^2-3x}=-1$ is not possible because the square root of a number is never a negative value.
So: $\sqrt{x^2-3x}=2$
$\sqrt{x^2-3x}=2$
$x^2-3x=2^2$
$x^2-3x=4$
$x^2-3x-4=0$
$(x-4)(x+1)=0$
$x=4$ or $x=-1$