what values of b satisfy 3(2b + 3)^2 = 36?\no b = (-3 + 2√3)/2 and (-3 - 2√3)/2\no b = (-3 + 2√3)/3 and (-3…

Explanation:

Step1: Isolate the squared - term

Divide both sides of the equation $3(2b + 3)^2=36$ by 3. $\frac{3(2b + 3)^2}{3}=\frac{36}{3}$, which simplifies to $(2b + 3)^2 = 12$.

Step2: Take the square - root of both sides

$2b+3=\pm\sqrt{12}=\pm2\sqrt{3}$.

Step3: Solve for b

First, consider the case when $2b + 3=2\sqrt{3}$. Then $2b=2\sqrt{3}-3$, and $b=\frac{-3 + 2\sqrt{3}}{2}$. Second, consider the case when $2b + 3=-2\sqrt{3}$. Then $2b=-2\sqrt{3}-3$, and $b=\frac{-3-2\sqrt{3}}{2}$.

Answer:

$b=\frac{-3 + 2\sqrt{3}}{2}$ and $b=\frac{-3-2\sqrt{3}}{2}$