# How do you solve 2x^2+5x-1=0 by completing the square?

Apr 19, 2018

$2 {\left(x + 1.25\right)}^{2} - 4.125 = 0$

#### Explanation:

We first take the first two terms and factor out the coefficient of ${x}^{2}$:
$\frac{2 {x}^{2}}{2} + \frac{5 x}{2} = 2 \left({x}^{2} + 2.5 x\right)$

Then we divide by $x$, half the integer and square what remains:
$2 \left({x}^{2} / x + 2.5 \frac{x}{x}\right) 2 = 2 \left(x + 2.5\right)$
$2 \left(x + 2 , \frac{5}{2}\right) = 2 \left(x + 1.25\right)$
$2 {\left(x + 1. .25\right)}^{2}$

Expand out the bracket:
$2 {x}^{2} + 2.5 x + 2.5 x + 2 \left({1.25}^{2}\right) = 2 {x}^{2} + 5 x + 3.125$

Make it equal the original equations:
$2 {x}^{2} + 5 x + 3.125 + a = 2 {x}^{2} + 5 x - 1$

Rearrange to find $a$:
$a = - 1 - 3.125 = - 4.125$

Put in $a$ to the factorised equation:
$2 {\left(x + 1.25\right)}^{2} - 4.125 = 0$