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

1 Answer
Jul 5, 2015

Answer:

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]#.