# How do you solve using the completing the square method 2x^2+8x-10=0?

Feb 3, 2017

$x = - 5$
$x = 1$

#### Explanation:

$2 {x}^{2} + 8 x - 10 = 2 \left[{x}^{2} + 4 x - 5\right]$

$2 \left[{x}^{2} + 4 x - 5\right] = 2 \left[{\left(x + 2\right)}^{2} - {2}^{2} - 5\right] = 2 \left[{\left(x + 2\right)}^{2} - 9\right] =$

$2 {\left(x + 2\right)}^{2} - 18 = 0$

$2 {\left(x + 2\right)}^{2} = 18$

${\left(x + 2\right)}^{2} = 9$

$x + 2 = \pm 3$

$x = \pm 3 - 2$

$x = - 5$

$x = 1$

Feb 3, 2017

$x = 1$ or $x = - 5$

#### Explanation:

Divide both sides by $2$ to get rid of the coefficient of ${x}^{2}$

${x}^{2} + 4 x - 5 = 0$

Add $5$ to both sides

${x}^{2} + 4 x = 5$

Take half the coefficient of $x$, square it, and add to both sides

• The coefficient of $x$ is $4$
• Half that is $2$
• The square of $2$ is $4$
• Adding $4$ to both sides gives

${x}^{2} + 4 x + 4 = 9$

Factoring the left hand side gives

${\left(x + 2\right)}^{2} = 9$

The square root of both sides gives

$x + 2 = 3$ or $x + 2 = - 3$
$x = 1$ or $x = - 5$