# How do you solve 2x^2 - 7x - 3 = 0 ?

Sep 9, 2015

$x = \frac{7 \pm \sqrt{73}}{4}$

#### Explanation:

An examination of the factors of $2$ and $- 3$ reveals that there are no integer solutions.

Therefore, we will use the quadratic root formula
$\textcolor{w h i t e}{\text{XXX}} x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
$\textcolor{w h i t e}{\text{XXXXXXXXXXX}}$for a quadratic in the general form $a {x}^{2} + b x + c = 0$

In this case
$\textcolor{w h i t e}{\text{XXX}} x = \frac{7 \pm \sqrt{{7}^{2} - 4 \left(2\right) \left(- 3\right)}}{2 \left(2\right)}$

$\textcolor{w h i t e}{\text{XXXX}} = \frac{7 \pm \sqrt{49 + 24}}{4}$

$\textcolor{w h i t e}{\text{XXXX}} = \frac{7 \pm \sqrt{73}}{4}$