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

${x}_{1} = \frac{1 + \sqrt{41}}{4}$
${x}_{2} = \frac{1 - \sqrt{41}}{4}$

#### Explanation:

From the given $2 {x}^{2} - x - 5 = 0$

Factor out the 2 from the first two terms so that the coefficient of ${x}^{2}$ is 1

$2 \left({x}^{2} - \frac{1}{2} x\right) - 5 = 0$

Now take note of the 1/2 coefficient of x. Divide this 1/2 by 2 it becomes 1/4. Square it, it will become 1/16. This 1/16 will be added and subtracted to the terms inside the grouping symbol.

Let us continue

$2 \left({x}^{2} - \frac{1}{2} x\right) - 5 = 0$

$2 \left({x}^{2} - \frac{1}{2} x + \frac{1}{16} - \frac{1}{16}\right) - 5 = 0$

Notice ${x}^{2} - \frac{1}{2} x + \frac{1}{16}$ is a PST-Perfect Square Trinomial and
${x}^{2} - \frac{1}{2} x + \frac{1}{16} = {\left(x - \frac{1}{4}\right)}^{2}$

so that

$2 \left(\left({x}^{2} - \frac{1}{2} x + \frac{1}{16}\right) - \frac{1}{16}\right) - 5 = 0$

$2 \left({\left(x - \frac{1}{4}\right)}^{2} - \frac{1}{16}\right) - 5 = 0$

Now, transpose the 5 to the right side then divide both sides by 2 then transpose the 1/16

$2 \left({\left(x - \frac{1}{4}\right)}^{2} - \frac{1}{16}\right) - 5 = 0$

$2 \left({\left(x - \frac{1}{4}\right)}^{2} - \frac{1}{16}\right) = 5$

$\frac{\cancel{2} \left({\left(x - \frac{1}{4}\right)}^{2} - \frac{1}{16}\right)}{\cancel{2}} = \frac{5}{2}$

${\left(x - \frac{1}{4}\right)}^{2} - \frac{1}{16} = \frac{5}{2}$

${\left(x - \frac{1}{4}\right)}^{2} = \frac{5}{2} + \frac{1}{16}$

Simplify

${\left(x - \frac{1}{4}\right)}^{2} = \frac{41}{16}$

Extract the square root of both sides

$\sqrt{{\left(x - \frac{1}{4}\right)}^{2}} = \pm \sqrt{\frac{41}{16}}$

$x - \frac{1}{4} = \pm \frac{1}{4} \sqrt{41}$

$x = \frac{1}{4} \pm \frac{1}{4} \sqrt{41}$

$x = \frac{1 \pm \sqrt{41}}{4}$

${x}_{1} = \frac{1 + \sqrt{41}}{4}$
${x}_{2} = \frac{1 - \sqrt{41}}{4}$

God bless...I hope the explanation is useful.