Question

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

Answer 1

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

Explanation:

`n^2-4n-12=0`

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

Second, add half the square of -4 (the coefficient of `n`) to both sides:
`n^2-4n+(1/2 * -4)^2=12+(1/2 * -4)^2`
`n^2-4n+(-2)^2=12+(-2)^2`
`n^2-4n+4=12+4`
`n^2-4n+4=16`

This made the left a perfect square, so change it:
`(n-2)^2`=16 then simplify

`n-2=+-sqrt(16)`
`n-2=+-4`
`n=2+-4`

The solution set is `[-2, 6]`.