How do you solve #x^2 = 2x + 15#?

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Answer 1 · 2016-04-09T12:12:38

#x=-3# or #x=5#

#x^2=2x+15#

#hArrx^2-2x-15=0#

Splitting the middle term in #-5x# and #3x# we get

#x^2-5x+3x-15=0# or

#x(x-5)+3(x-5)=0# or

#(x+3)(x-5)=0#

Hence #x=-3# or #x=5#