# How do you solve 3x^2 + 20x + 36 = 4 by completing the square?

$f \left(x\right) = 3 {x}^{2} + 20 x + 32$= $3 \left({x}^{2} + \frac{20 x}{3} + \frac{32}{3}\right) = 0$
${x}^{2} + \frac{20 x}{3} + \frac{100}{9} - \frac{100}{9} + \frac{32}{3} = 0$
$= {\left(x + \frac{10}{3}\right)}^{2} - \frac{4}{9} = 0$ ->$\left(x + \frac{10}{3}\right) = \pm \frac{2}{3}$
$\left(x + \frac{10}{3}\right) = \frac{2}{3} \to x = - \frac{10}{3} + \frac{2}{3} = - \frac{8}{3}$
$\left(x + \frac{10}{3}\right) = - \frac{2}{3} \to x = - \frac{10}{3} - \frac{2}{3} = - 12 / 3 = - 4$