# How do you find the zeros, real and imaginary, of y=8x^2-4x-11 using the quadratic formula?

Feb 19, 2016

#### Answer:

$x = \frac{1 \pm \sqrt{23}}{4}$

#### Explanation:

For $a {x}^{2} + b x + c = 0$:

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

Therefore $a = 8 , b = - 4 , c = - 11$ and:

$x = \frac{4 \pm \sqrt{{\left(- 4\right)}^{2} - 4 \cdot 8 \cdot - 11}}{2 \cdot 8}$
$x = \frac{4 \pm \sqrt{16 + 352}}{16}$
$x = \frac{4 \pm \sqrt{368}}{16}$
$x = \frac{4 \pm 4 \sqrt{23}}{16}$
$x = \frac{1 \pm \sqrt{23}}{4}$