# How do you solve x^2+18x=10 by completing the square?

Jun 7, 2018

$x = - 9 \pm \sqrt{91}$

#### Explanation:

For completing the square, you take half the coefficient of the x value, square it, and add it. So in this case, 81 I suppose?
I apologize, I don't quite remember how to do this. However there is already a 10...

Oh! I got it!

So add 91 and then you would get:

${x}^{2} + 18 x + 81 = 91$

Then you factor:

${\left(x + 9\right)}^{2} = 91$

Square root:

$x + 9 = \pm \sqrt{91}$

Solve by subtracting 9:

$x = - 9 \pm \sqrt{91}$

Jun 7, 2018

$x = - 9 \pm \sqrt{91}$

#### Explanation:

$a {x}^{2} + b x + c$

To complete the square a must be 1 (which it is) and:

$c = {\left(\frac{b}{2}\right)}^{2}$

${x}^{2} + 18 x = 10$

${x}^{2} + 18 x + c = 10 + c$

notice we have to add c to both sides of the equation so we don't alter the value.

$c = {\left(\frac{18}{2}\right)}^{2} = 81$

${x}^{2} + 18 x + 81 = 10 + 81$

now factor the left side:

$\left(x + 9\right) \left(x + 9\right) = 91$

${\left(x + 9\right)}^{2} = 91$

now we solve:

$\sqrt{{\left(x + 9\right)}^{2}} = \pm \sqrt{91}$

$x + 9 = \pm \sqrt{91}$

$x = - 9 \pm \sqrt{91}$