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

Jan 23, 2016

$x = \frac{1 \pm \sqrt{37}}{2}$

#### Explanation:

The quadratic formula find the zeroes of a quadratic equation in the form $y = a {x}^{2} + b x + c$ through

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

In the given quadratic equation, we know that $a = 1 , b = - 1 , c = - 9$. Plugging these into the quadratic formula gives

$x = \frac{- \left(- 1\right) \pm \sqrt{{\left(- 1\right)}^{2} - \left(4 \times - 9 \times 1\right)}}{2 \times 1}$

$x = \frac{1 \pm \sqrt{1 - \left(- 36\right)}}{2}$

$x = \frac{1 \pm \sqrt{37}}{2}$

Thus, the two times when the function equal $0$ are at $x = \frac{1 + \sqrt{37}}{2}$ and $x = \frac{1 - \sqrt{37}}{2}$.