# How do you find the zeros, real and imaginary, of y=-x^2 -2x +5 using the quadratic formula?

Nov 5, 2017

The quadratic formula says that the zeros of a quadratic function can be calculated as:

and

where

## $\Delta = {b}^{2} - 4 a c$

Here we have:

$\Delta = {\left(- 2\right)}^{2} - 4 \cdot \left(- 1\right) \cdot 5 = 4 + 20 = 24$

$\sqrt{\Delta} = \sqrt{24} = 2 \sqrt{6}$

The discriminant $\left(\Delta\right)$ is greater than zero, so the function has 2 different real zeros:

${x}_{1} = \frac{2 - 2 \sqrt{6}}{- 2} = - 1 + \sqrt{6}$

${x}_{1} = \frac{2 + 2 \sqrt{6}}{- 2} = - 1 - \sqrt{6}$