# How do you solve 2x^2-9=0 using the quadratic formula?

May 11, 2016

$x = \frac{\sqrt{72}}{4}$
$x = - \left(\frac{\sqrt{72}}{4}\right)$

#### Explanation:

Given -

$2 {x}^{2} - 9 = 0$

Quadratic equations normally looks like this

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

In the given equation the $b x$ term is missing

We shall supply it

$2 {x}^{2} + 0 x - 9$

Then as per formula the roots are

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

We shall substitute the values in the formula

$x = \frac{\left(- 0\right) \pm \sqrt{{0}^{2} - \left(4 \times 2 \times \left(- 9\right)\right)}}{2 \times 2}$

$x = \frac{\pm \sqrt{0 - \left(- 72\right)}}{4}$

The two roots are

$x = \frac{\sqrt{72}}{4}$
$x = - \left(\frac{\sqrt{72}}{4}\right)$