# How do you solve y^2 +6y = 9 using the quadratic formula?

Nov 8, 2015

$y = - 3 \left(1 - \sqrt{2}\right) , - 3 \left(1 + \sqrt{2}\right)$

#### Explanation:

Move all terms to the left side.

${y}^{2} + 6 y - 9 = 0$ is a quadratic equation, $a {x}^{2} + b x + c$, where $a = 1 , b = 6 , \mathmr{and} c = - 9$.

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Substitute $y$ for $x$.

$y = \frac{- 6 \pm \sqrt{{6}^{2} - 4 \cdot 1 \cdot - 9}}{2 \cdot 1}$

Simplify.

$y = \frac{- 6 \pm \sqrt{36 + 36}}{2} =$

$y = \frac{- 6 \pm \sqrt{72}}{2}$

Factor $\sqrt{72}$.

$y = \frac{- 6 \pm \sqrt{2 \times 2 \times 2 \times 3 \times 3}}{2} =$

$y = \frac{- 6 \pm \sqrt{2 \times {2}^{2} \times {3}^{2}}}{2} =$

$y = \frac{- 6 \pm 2 \times 3 \sqrt{2}}{2} =$

$y = \frac{- 6 \pm 6 \sqrt{2}}{2} =$

$y = - 3 \pm 3 \sqrt{2}$

$y = - 3 \left(1 - \sqrt{2}\right) , - 3 \left(1 + \sqrt{2}\right)$