Answer:
[tex]\frac{-3+ 2\sqrt{3}}{2}\textrm{ and }\frac{-3- 2\sqrt{3}}{2}[/tex]
Step-by-step explanation:
[tex]3(2b+3)^{2}=36\\\frac{3(2b+3)^{2}}{3}=\frac{36}{3}\\(2b+3)^{2}=12[/tex]
Taking square root both sides, we get
[tex]\sqrt{(2b+3)^{2}}=\pm \sqrt{12}\\2b+3=\pm \sqrt{4\times 3}\\2b+3=\pm 2\sqrt{3}\\2b+3-3=\pm 2\sqrt{3}-3\\2b=-3\pm 2\sqrt{3}\\b=\frac{-3\pm 2\sqrt{3}}{2}\\b=\frac{-3+ 2\sqrt{3}}{2}\textrm{ or }b=\frac{-3- 2\sqrt{3}}{2}[/tex]
Therefore, the values of [tex]b[/tex] are:
[tex]\frac{-3+ 2\sqrt{3}}{2}\textrm{ and }\frac{-3- 2\sqrt{3}}{2}[/tex]
