We are given the equation:

$4(3b + 2)^2 = 64$

To solve for $b$, let's break it down step by step:

Step 1: Simplify the equation

Divide both sides of the equation by 4:

$(3b + 2)^2 = \frac{64}{4} = 16$

Step 2: Take the square root of both sides

Now, take the square root of both sides of the equation:

$3b + 2 = \pm \sqrt{16}$

$3b + 2 = \pm 4$

Step 3: Solve for $b$

We now have two cases to solve:

1. $3b + 2 = 4$
2. $3b + 2 = -4$

Case 1: $3b + 2 = 4$

Subtract 2 from both sides:

$3b = 4 - 2 = 2$

Now, divide by 3:

$b = \frac{2}{3}$

Case 2: $3b + 2 = -4$

Subtract 2 from both sides:

$3b = -4 - 2 = -6$

Now, divide by 3:

$b = \frac{-6}{3} = -2$

Final Answer:

The values of $b$ are $b = \frac{2}{3}$ and $b = -2$.

Let me know if you would like any further details or clarifications!

Here are some related questions:

1. How do you solve quadratic equations in general?
2. What happens when you take the square root of a negative number?
3. How do you check if your solutions for $b$ are correct?
4. Can there be more than two solutions to this type of equation?
5. How does squaring affect the outcome of equations like these?

Tip: When solving equations involving squares, always remember to consider both the positive and negative square roots!