# How do you solve x^2 + 6x - 9 = 0 by completing the square?

Sep 11, 2017

x = -3 ± 3sqrt2

#### Explanation:

For a polynomial ${x}^{2} + b x + c$, the completed square is of the form ${\left(x + \frac{b}{2}\right)}^{2} - {\left(\frac{b}{2}\right)}^{2} + c$

$\therefore {x}^{2} + 6 x - 9 = {\left(x + 3\right)}^{2} - 9 - 9 = {\left(x + 3\right)}^{2} - 18$

${\left(x + 3\right)}^{2} - 18 = 0$

${\left(x + 3\right)}^{2} = 18$

x+3 = ±sqrt18 = ±3sqrt2

x = -3 ± 3sqrt2