# How do you find the roots, real and imaginary, of y= 5x^2 - 2x-45  using the quadratic formula?

Jan 30, 2016

Since ${b}^{2} - 4 a c$ is 904 which is a positive number, the given quadratic equation has real roots.

$x = \frac{2 \pm \sqrt{904}}{10}$

#### Explanation:

When the quadratic equation is in the form -

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

Then its roots are given by

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

In this ${b}^{2} - 4 a c$ determines whether a given equation has real roots or imaginary roots.

If ${b}^{2} - 4 a c$ is positive the roots are positive.

If ${b}^{2} - 4 a c$ is negative the roots are negative.

In our case -

${\left(- 2\right)}^{2} - \left(4 \times 5 \times - 45\right)$
$4 - \left(- 900\right)$
$4 + 900 = 904$

Since ${b}^{2} - 4 a c$ is 904 which is a positive number, the given quadratic equation has real roots. So we should be able to plug into the quadratic formula without having imaginary numbers:

$x = \frac{2 \pm \sqrt{904}}{2 \left(5\right)}$

$x = \frac{2 \pm \sqrt{904}}{10}$