How do you solve #x^2+x-8=4#?

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Answer 1 · 2016-02-10T20:51:14

#x^2 + x - 8 - 4 = 0#
#x^2 + x - 12 = 0#
#x^2 +4 x - 3x -12 = 0#
#x(x+4) - 3(x+4) =0#
#(x+4)*(x-3) =0#
#x+4=0, x-3=0#
#x=-4, x=3 #