# How do you find the zeros, real and imaginary, of  y=5(x-3)^2-56  using the quadratic formula?

Jan 10, 2016

$x = 3 \pm 3 , 35$

#### Explanation:

Develop $y = 5 \left({x}^{2} - 6 x + 9\right) - 56 = 5 {x}^{2} - 30 x + 45 - 56$
$y = 5 {x}^{2} - 30 x - 11 = 0$
$D = {d}^{2} = {b}^{2} - 4 a c = 900 + 220 = 1120$ --> $d = \pm 33.47$.
There are 2 real roots:
$x = \frac{30}{10} \pm \frac{33.47}{10} = 3 \pm 3.35$
x1 = 6.35 and x2 = 0.35