the correct question is
What values of b satisfy 3(2b+3)^2 = 36
we have
\(3(2b+3)^2 = 36\)
Divide both sides by \(3\)
\((2b+3)^2 = 12\)
take the square root of both sides
\(( 2b+3)} =(+ /-) \sqrt{12} \\ 2b=(+ /-) \sqrt{12}-3\)
\(b1=\frac{\sqrt{12}}{2} -\frac{3}{2}\)
\(b1=\sqrt{3} -\frac{3}{2}\)
\(b2=\frac{-\sqrt{12}}{2} -\frac{3}{2}\)
\(b2=-\sqrt{3} -\frac{3}{2}\)
therefore
the answer is
the values of b are
\(b1=\sqrt{3} -\frac{3}{2}\)
\(b2=-\sqrt{3} -\frac{3}{2}\)
