# How do you find the zeros of  y = x^2 - 5x -13  using the quadratic formula?

Mar 13, 2018

The quadratic formula always gives two answers. In this case, $x = 6.8875$ and $x = - 1.8875$

#### Explanation:

To find the 'zeros', you want to solve the function for when y=0:

$0 = {x}^{2} - 5 x - 13$

There is a general form for this, which is what we will be using to describe the quadratic formula:

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

The quadratic formula looks like this:

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

For this case:
$a = 1$
$b = - 5$
$c = - 13$

Now, let's plug those into the formula:

$x = \frac{- \left(- 5\right) \pm \sqrt{{\left(- 5\right)}^{2} - 4 \left(1\right) \left(- 13\right)}}{2 \left(1\right)}$

$x = \frac{5 \pm \sqrt{25 - 4 \cdot \left(- 13\right)}}{2}$

$x = \frac{5 \pm \sqrt{25 + 52}}{2} \Rightarrow x = \frac{5 \pm \sqrt{77}}{2}$

Now, we solve for both scenarios of x:

color(red)(x=(5+sqrt(77))/2 rArr x=6.8875

color(blue)(x=(5-sqrt(77))/2 rArr x=-1.8875