# How do you solve by completing the square for 2x^2+10x-4=0?

Mar 13, 2018

$2 {\left(x + \frac{5}{2}\right)}^{2} - \frac{33}{2}$

#### Explanation:

Take 2 out of the equation:
$2 \left({x}^{2} + 5 x - 2\right) = 0$

Complete the square in the brackets:
$2 \left({\left(x + \frac{5}{2}\right)}^{2} - \frac{25}{4} - 2\right)$

Simplify the equation:
$2 \left({\left(x + \frac{5}{2}\right)}^{2} - \frac{33}{4}\right)$
$2 {\left(x + \frac{5}{2}\right)}^{2} - \frac{33}{2}$

Mar 13, 2018

$2 {x}^{2} + 10 x - 4 = 0$

First divide by $2$ so that we have $1 {x}^{2} \text{ } a = 1$

${x}^{2} + 5 x - 2 = 0$

${x}^{2} + 5 x \text{ "=2" } \leftarrow$ move the constant to the RHS

${x}^{2} + 5 x + {\left(\frac{5}{2}\right)}^{2} = 2 + {\left(\frac{5}{2}\right)}^{2} \text{ } \leftarrow$ add ${\left(\frac{b}{2}\right)}^{2}$ to both sides.

By this process you have written the left side as a 'perfect square'
This step is the 'completing the square '- add in the missing term to create a square.

Write the left side as the square of a binomial.

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

$x + \frac{5}{2} = \pm \sqrt{\frac{9}{2}} \text{ } \leftarrow$ find the square root of both sides.

Find the two possible solutions:

$x = \frac{3}{+ \sqrt{2}} - 2.5 = - 0.379$ (3 dp)

$x = \frac{3}{- \sqrt{2}} - 2.5 = - 4.624$