Answer
$x = 3$ or $x = 4$
Work Step by Step
$$\sqrt {x^{2} - 3x} = 2x - 6$$
$(\sqrt {x^{2} -3x})^{2} = (2x - 6)^{2}$
$(x^{2} - 3x) = (4x^{2} - 24x + 36)$
$0 = 4x^{2} - x^{2} - 24x + 3x + 36$
$0 = 3x^{2} -21x +36$
$0 = 3(x^{2} - 7x + 12)$
$0 = x^{2} - 7x + 12$
since two factors of 12 that, when added, give 7, are $3$ and $3$, we can factorize the equation in the following manner:
$0 = (x - 3)(x - 4)$
where:
$(x - 3) = 0$ or $(x - 4) = 0$
$x = 3$ or $x = 4$
