SOLUTION: {{{ (3x+1)^2-36=0 }}}

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Question 283164: +%283x%2B1%29%5E2-36=0+
Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
This a difference of squares.
Consider (3x+1) as a and 6 as b
since 6^2=36
a^2-b^2=(a-b)*(a+b)
(3x+1-6)*(3x+1+6)=0
add the numbers
set each factor to zero
and solve