# How do you find the zeros, real and imaginary, of  y=2(x+4)^2+4  using the quadratic formula?

May 21, 2018

$x = - 4 \pm i \sqrt{2}$

#### Explanation:

2 Ways to Solve:

Solve by Square Roots:
$2 {\left(x + 4\right)}^{2} + 4 = 0$
$2 {\left(x + 4\right)}^{2} = - 4$
${\left(x + 4\right)}^{2} = - 2$
$\left(x + 4\right) = \pm \sqrt{- 2}$
$x = - 4 \pm i \sqrt{2}$

Need to multiply out first
$y = 2 {\left(x + 4\right)}^{2} + 4$
$y = 2 \left(x + 4\right) \left(x + 4\right) + 4$
$y = 2 \left({x}^{2} + 8 x + 16\right) + 4$
$y = 2 {x}^{2} + 16 x + 32 + 4$
$y = 2 {x}^{2} + 16 x + 36$

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
$x = \frac{- \left(16\right) \pm \sqrt{{\left(16\right)}^{2} - 4 \left(2\right) \left(36\right)}}{2 \left(2\right)}$
$x = \frac{- 16 \pm \sqrt{256 - 288}}{4}$
$x = \frac{- 16 \pm \sqrt{- 32}}{4}$
$x = \frac{- 16 \pm 4 i \sqrt{2}}{4}$
$x = - 4 \pm i \sqrt{2}$