# How do you solve x^2-2x-14=0 by completing the square?

Jul 27, 2017

$x = \pm \sqrt{15} + 1$

#### Explanation:

First, you have to move the 14 to the other side by adding it :

${x}^{2} - 2 x = 14$

To complete the square you divide the term with one x by 2 and then take that number and square it :

$\frac{2}{2} = 1$ , ${1}^{2} = 1$

We then take that 1 and add it to both sides of the equation :

${x}^{2} - 2 x + 1 = 15$

Now we can factor the left side of the equation :

${\left(x - 1\right)}^{2} = 15$

Take the square root of both sides to get rid of the square on the left hand side (be sure to add the $\pm$ on the 15) :

$\sqrt{{\left(x - 1\right)}^{2}} = \pm \sqrt{15}$

which then equals...

$x - 1 = \pm \sqrt{15}$

Now get x by itself :

$x = \pm \sqrt{15} + 1$