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

Jul 8, 2015

Force a perfect square trinomial on the left side. Take the square root of both sides and solve for$x$.

#### Explanation:

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

Add $2$ to both sides of the equation.

${x}^{2} + 10 x = 2$

Force a perfect square trinomial on the left side of the equation.

Divide the coefficient of the $x$ term and square the result. Add it to both sides of the equation.

${\left(\frac{10}{2}\right)}^{2} = {5}^{2} = 25$

${x}^{2} + 10 x + 25 = 2 + 25$ =

${x}^{2} + 10 x + 25 = 27$

We now have a perfect square trinomial on the left side with the form ${a}^{2} + 2 a b + {b}^{2} = {\left(a + b\right)}^{2}$.

$a = x ,$ $b = 5$

${\left(x + 5\right)}^{2} = 27$

Take the square root of both sides.

$\left(x + 5\right) = \pm \sqrt{27}$ =

$x + 5 = \pm \sqrt{3} \sqrt{9}$ =

$x + 5 = \pm 3 \sqrt{3}$ =

Subtract $5$ from both sides.

$x = - 5 \pm 3 \sqrt{3}$

$x = - 5 + 3 \sqrt{3}$

$x = - 5 - 3 \sqrt{3}$

$x = - 5 + 3 \sqrt{3} , - 5 - 3 \sqrt{3}$