How do you solve #x^2+2x-3=0# by completing the square?

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Answer 1 · 2015-05-05T11:22:22

#x^2+2x-3=0#

If #x^2+2x+a^2# is a square then #a^2=1#

#x^2+2x+1 -1 -3 = 0#

#(x+1)^2 = 4#

#x+1 = +-sqrt(4)#

#x = -1 +-2#

#x=-3 " or " x=1#