Solve for b 4(3b+2)^2=64

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

Divide each term in $4 {(3 b + 2)}^{2} = 64$ by $4$ and simplify.

Divide each term in $4 {(3 b + 2)}^{2} = 64$ by $4$ .

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

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

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

Divide ${(3 b + 2)}^{2}$ by $1$ .

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

Simplify the right side.

Divide $64$ by $4$ .

${(3 b + 2)}^{2} = 16$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

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

Simplify $\pm \sqrt{16}$ .

Rewrite $16$ as $4^{2}$ .

$3 b + 2 = \pm \sqrt{4^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$3 b + 2 = \pm 4$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$3 b + 2 = 4$

Move all terms not containing $b$ to the right side of the equation.

Subtract $2$ from both sides of the equation.

$3 b = 4 - 2$

Subtract $2$ from $4$ .

$3 b = 2$

Divide each term in $3 b = 2$ by $3$ and simplify.

Divide each term in $3 b = 2$ by $3$ .

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

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $b$ by $1$ .

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

Next, use the negative value of the $\pm$ to find the second solution.

$3 b + 2 = - 4$

Move all terms not containing $b$ to the right side of the equation.

Subtract $2$ from both sides of the equation.

$3 b = - 4 - 2$

Subtract $2$ from $- 4$ .

$3 b = - 6$

Divide each term in $3 b = - 6$ by $3$ and simplify.

Divide each term in $3 b = - 6$ by $3$ .

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

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $b$ by $1$ .

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

Simplify the right side.

Divide $- 6$ by $3$ .

$b = - 2$

The complete solution is the result of both the positive and negative portions of the solution.

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

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$b = 0.\overline{6}, - 2$

Solve for b 4(3b+2)^2=64

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