# How do you solve x^2-5x-3/4=0 by completing the square?

May 13, 2016

$x = \frac{5}{2} \pm \sqrt{7}$

#### Explanation:

Premultiply by $4$ first to cut down on the fractions involved:

$0 = 4 \left({x}^{2} - 5 x - \frac{3}{4}\right)$

$= 4 {x}^{2} - 20 x - 3$

$= {\left(2 x\right)}^{2} - 2 \left(2 x\right) \left(5\right) - 3$

$= {\left(2 x - 5\right)}^{2} - {5}^{2} - 3$

$= {\left(2 x - 5\right)}^{2} - 28$

$= {\left(2 x - 5\right)}^{2} - {2}^{2} \cdot 7$

$= {\left(2 x - 5\right)}^{2} - {\left(2 \sqrt{7}\right)}^{2}$

$= \left(\left(2 x - 5\right) - 2 \sqrt{7}\right) \left(\left(2 x - 5\right) + 2 \sqrt{7}\right)$

$= \left(2 x - 5 - 2 \sqrt{7}\right) \left(2 x - 5 + 2 \sqrt{7}\right)$

$= 4 \left(x - \frac{5}{2} - \sqrt{7}\right) \left(x - \frac{5}{2} + \sqrt{7}\right)$

So $x = \frac{5}{2} \pm \sqrt{7}$