Let's solve the equation \( 3(2b + 3)^2 = 36 \).
First, we can simplify the equation by dividing both sides by 3:
\[ (2b + 3)^2 = \frac{36}{3} \]
This simplifies to:
\[ (2b + 3)^2 = 12 \]
Next, we take the square root of both sides:
\[ 2b + 3 = \sqrt{12} \quad \text{or} \quad 2b + 3 = -\sqrt{12} \]
Since \(\sqrt{12} = 2\sqrt{3}\), we can rewrite the equations as:
- \(2b + 3 = 2\sqrt{3}\)
- \(2b + 3 = -2\sqrt{3}\)
Now, we will solve for \(b\) in both cases.
Solving the first equation:
\[ 2b + 3 = 2\sqrt{3} \] Subtract 3 from both sides:
\[ 2b = 2\sqrt{3} - 3 \] Divide by 2:
\[ b = \frac{2\sqrt{3} - 3}{2} \]
Solving the second equation:
\[ 2b + 3 = -2\sqrt{3} \] Subtract 3 from both sides:
\[ 2b = -2\sqrt{3} - 3 \] Divide by 2:
\[ b = \frac{-2\sqrt{3} - 3}{2} \]
Thus, the solutions for \(b\) are:
\[ b = \frac{2\sqrt{3} - 3}{2} \quad \text{and} \quad b = \frac{-2\sqrt{3} - 3}{2} \]
These solutions correspond to the values provided in the options given. To match these expressions to a simplified format, we can write them as:
- \(b = \frac{-3 + 2\sqrt{3}}{2}\)
- \(b = \frac{-3 - 2\sqrt{3}}{2}\)
These match to the first suggested option:
\[ b = \frac{-3 + 2\sqrt{3}}{2} \text{ and } b = \frac{-3 - 2\sqrt{3}}{2} \]
Therefore, the final answer that satisfies the equation is:
\(\boxed{b = \frac{-3 + 2\sqrt{3}}{2} \text{ and } b = \frac{-3 - 2\sqrt{3}}{2}}\)