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

Apr 19, 2016

Look for the steps below.

#### Explanation:

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

Bring the constant to the other side of the equal sign.
${x}^{2} - 10 x = 0 + 4$

Add the square of half of the b term and evaluate.
${x}^{2} - 10 x + {\left(- \frac{10}{2}\right)}^{2} = 4 + {\left(\frac{- 10}{2}\right)}^{2}$
${x}^{2} - 10 x + {\left(- 5\right)}^{2} = 4 + {\left(- 5\right)}^{2}$
${x}^{2} - 10 x + 25 = 4 + 25$
${x}^{2} - 10 x + 25 = 29$

Make the left side of the equal sign as a perfect square binomial by having x and then the half of the b term.
${\left(x - 5\right)}^{2} = 29$

To find $x$ you need to make sure that it is out of the square, so you need to find the square root of $x - 5$ and do the same to 29.

$\sqrt{{\left(x - 5\right)}^{2}} = \sqrt{29}$
$x - 5 = \sqrt{29}$

To find x you need to isolate it.
Answer: $x = 5 \pm \sqrt{29}$

The reason for why I put the $\pm$ sign is because the square root of 29 can be both positive and negative. You would need to find both of the results from adding and subtracting to figure out the root of the quadratic.