How do you solve #(x + 4) ( x + 2) = 15#?

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Answer 1 · 2016-09-15T03:29:50

#x=1# or #x=-7#

#(x+4)(x+2)=15#

#hArrx(x+2)+4(x+2)=15# or

#x^2+2x+4x+8=15# or

#x^2+6x-7=0# or

#x^2+7x-x-7=0# or

#x(x+7)-1(x+7)=0# or

#(x-1)(x+7)=0#

i.e. either #x-1=0# or #x+7=0#.

Hence #x=1# or #x=-7#