# How do you find the roots, real and imaginary, of y= 4x^2-36  using the quadratic formula?

Dec 17, 2015

$x = \pm 3$

#### Explanation:

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

for the quadratic equation in the general form

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

Your equation does not seem to fit this mold, but it can be rewritten as

$y = 4 {x}^{2} + 0 x - 36$

So,

$a = 4$
$b = 0$
$c = - 36$

Plug these into the formula.

$x = \frac{- 0 \pm \sqrt{{0}^{2} - 4 \left(4\right) \left(- 36\right)}}{2 \left(4\right)}$

$x = \frac{\pm \sqrt{576}}{8}$

$x = \frac{\pm 24}{8}$

$x = \pm 3$

These are the two real roots.