# How do you solve  3(x-5)^2 - 12 = 0?

$x = 5 + 2 = 7 \mathmr{and} x = 5 - 2 = 3$
$3 {\left(x - 5\right)}^{2} = 12$
$\implies {\left(x - 5\right)}^{2} = \frac{12}{3} = 4$
$\implies \left(x - 5\right) = \pm \sqrt{4} = \pm 2$
$x = 5 + 2 = 7 \mathmr{and} x = 5 - 2 = 3$