# How do you find the zeros, real and imaginary, of y=x^2+3x-9 using the quadratic formula?

Aug 3, 2018

The zeros are $x = \frac{- 3 + 3 \sqrt{5}}{2}$ and $x = \frac{- 3 - 3 \sqrt{5}}{2}$.

#### Explanation:

$y = {x}^{2} + 3 x - 9$

The quadratic formula is $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$, where from our equation we know that $a = 1 , b = 3 , c = - 9$.

Plug them into the formula:
$x = \frac{- 3 \pm \sqrt{{3}^{2} - 4 \left(1\right) \left(- 9\right)}}{2 \left(1\right)}$

$x = \frac{- 3 \pm \sqrt{9 + 36}}{2}$

$x = \frac{- 3 \pm \sqrt{45}}{2}$

$x = \frac{- 3 \pm 3 \sqrt{5}}{2}$

Hope this helps!