Solve for b 12=-6 square root of b+4

$12 = - 6 \sqrt{b + 4}$

Rewrite the equation as $- 6 \sqrt{b + 4} = 12$ .

$- 6 \sqrt{b + 4} = 12$

To remove the radical on the left side of the equation, square both sides of the equation.

${(- 6 \sqrt{b + 4})}^{2} = 12^{2}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{b + 4}$ as ${(b + 4)}^{\frac{1}{2}}$ .

${(- 6 {(b + 4)}^{\frac{1}{2}})}^{2} = 12^{2}$

Simplify the left side.

Simplify ${(- 6 {(b + 4)}^{\frac{1}{2}})}^{2}$ .

Apply the product rule to $- 6 {(b + 4)}^{\frac{1}{2}}$ .

${(- 6)}^{2} {({(b + 4)}^{\frac{1}{2}})}^{2} = 12^{2}$

Raise $- 6$ to the power of $2$ .

$36 {({(b + 4)}^{\frac{1}{2}})}^{2} = 12^{2}$

Multiply the exponents in ${({(b + 4)}^{\frac{1}{2}})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$36 {(b + 4)}^{\frac{1}{2} \cdot 2} = 12^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$36 {(b + 4)}^{\frac{1}{2} \cdot 2} = 12^{2}$

Rewrite the expression.

$36 {(b + 4)}^{1} = 12^{2}$

Simplify.

$36 (b + 4) = 12^{2}$

Apply the distributive property.

$36 b + 36 \cdot 4 = 12^{2}$

Multiply $36$ by $4$ .

$36 b + 144 = 12^{2}$

Simplify the right side.

Raise $12$ to the power of $2$ .

$36 b + 144 = 144$

Solve for $b$ .

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

Subtract $144$ from both sides of the equation.

$36 b = 144 - 144$

Subtract $144$ from $144$ .

$36 b = 0$

Divide each term in $36 b = 0$ by $36$ and simplify.

Divide each term in $36 b = 0$ by $36$ .

$\frac{36 b}{36} = \frac{0}{36}$

Simplify the left side.

Cancel the common factor of $36$ .

Cancel the common factor.

$\frac{36 b}{36} = \frac{0}{36}$

Divide $b$ by $1$ .

$b = \frac{0}{36}$

Simplify the right side.

Divide $0$ by $36$ .

$b = 0$

Exclude the solutions that do not make $12 = - 6 \sqrt{b + 4}$ true.

$No solution$

Solve for b 12=-6 square root of b+4

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