# How do you solve n^2-4n-12=0 by completing the square?

Jul 5, 2015

Move the 12 over, then add half the square of -4 to both sides and simplify.

#### Explanation:

${n}^{2} - 4 n - 12 = 0$

First, move the 12 over to the right hand side:
${n}^{2} - 4 n = 12$

Second, add half the square of -4 (the coefficient of $n$) to both sides:
${n}^{2} - 4 n + {\left(\frac{1}{2} \cdot - 4\right)}^{2} = 12 + {\left(\frac{1}{2} \cdot - 4\right)}^{2}$
${n}^{2} - 4 n + {\left(- 2\right)}^{2} = 12 + {\left(- 2\right)}^{2}$
${n}^{2} - 4 n + 4 = 12 + 4$
${n}^{2} - 4 n + 4 = 16$

This made the left a perfect square, so change it:
${\left(n - 2\right)}^{2}$=16 then simplify

$n - 2 = \pm \sqrt{16}$
$n - 2 = \pm 4$
$n = 2 \pm 4$

The solution set is $\left[- 2 , 6\right]$.