How do you solve #(2x + 1) ( x + 3) = 42#?

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Answer 1 · 2017-07-16T15:13:15

#x=3# or #x=-6 1/2#

#(2x+1)(x+3)=43#

We first expand the left side.

#2x(x+3)+1(x+3)=42#

#2x^2+6x+x+3=42#

#2x^2+7x+3=42#

Subtract #42# from each side.

#2x^2+7x+3-43=42-42#

#2x^2+7x-39=0#

Factorise.

#2x^2+13x-6x-39=0#

#x(2x+13)-3(2x+13)=0#

#(x-3)(2x+13)=0#

#x-3=0# or #2x+13=0#

#x=3# or #x=-13/2=-6 1/2#