# How do you solve using the completing the square method 3x^2 +18x +20 = 0?

Apr 2, 2016

Dividing both sides of the Eq by 3 we have
${x}^{2} + 6 x + \frac{20}{3} = 0$
$\implies {x}^{2} + 2 \times 3 \times x + {3}^{3} - {3}^{2} + \frac{20}{3} = 0$
$\implies {\left(x + 3\right)}^{2} - 9 + \frac{20}{3} = 0$
$\implies {\left(x + 3\right)}^{2} - \frac{7}{3} = 0$
$\implies {\left(x + 3\right)}^{2} - {\left(\sqrt{\frac{7}{3}}\right)}^{2} = 0$
$\implies \left(x + 3 + \sqrt{\frac{7}{3}}\right) \left(x + 3 - \sqrt{\frac{7}{3}}\right) = 0$
$\therefore$
$\left(x = - 3 - \sqrt{\frac{7}{3}}\right)$
and
$\left(x = - 3 + \sqrt{\frac{7}{3}}\right)$