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}
