what values of b satisfy 4(3b + 2)^2 = 64?\no b = 2/3 and b = -2\no b = 2 and b = 10/3\no b = 2/3 and b =…

Explanation:

Step1: Simplify the equation

Divide both sides of $4(3b + 2)^2=64$ by 4. We get $(3b + 2)^2 = 16$.

Step2: Take square - root of both sides

$3b+2=\pm\sqrt{16}=\pm4$.

Step3: Solve for $b$ when $3b + 2 = 4$

Subtract 2 from both sides: $3b=4 - 2=2$. Then divide by 3, so $b=\frac{2}{3}$.

Step4: Solve for $b$ when $3b + 2=-4$

Subtract 2 from both sides: $3b=-4 - 2=-6$. Then divide by 3, so $b=-2$.

Answer:

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