How do you solve #x^2 - 4x - 5 = 0# by completing the square?

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Answer 1 · 2018-05-31T00:56:34

Below

#x^2-4x-5=0#

#x^2-4x=5#

#x^2-4x+(4/2)^2=5+(4/2)^2#

#x^2-4x+4 = 5+4#

#(x-2)^2=9#

#x-2=+-3#

#x-2=3# OR #x-2=-3#

#x-2=3#
#x=5#

#x-2=-3#
#x=-1#