# How do you solve the quadratic equation by completing the square: y^2 + 16y = 2?

Jun 14, 2018

$y = - 8 \pm \sqrt{66}$

#### Explanation:

to complete the square you use the formula:

$a {x}^{2} + b x + c$

a must equal 1

$c = {\left(\frac{b}{2}\right)}^{2}$

the completed square is:

${\left(x + \frac{b}{2}\right)}^{2}$

Here we go, in your function the y is the general formula's x:

${y}^{2} + 16 y = 2$

${y}^{2} + 16 y + \underbrace{c = 2 + c}$
we add c to both sides so we don't alter the equation

now solve c:

$c = {\left(\frac{b}{2}\right)}^{2} = {\left(\frac{16}{2}\right)}^{2} = 64$

${y}^{2} + 16 y + 64 = 2 + 64$

now complete the square:

${\left(y + 8\right)}^{2} = 66$

Now solve:

$\sqrt{{\left(y + 8\right)}^{2}} = \pm \sqrt{66}$

$y + 8 = \pm \sqrt{66}$

$y = - 8 \pm \sqrt{66}$