Solve for b 2 square root of b+5=11- square root of b

$2 \sqrt{b} + 5 = 11 - \sqrt{b}$

Solve for $\sqrt{b}$ .

Move all terms containing $\sqrt{b}$ to the left side of the equation.

Add $\sqrt{b}$ to both sides of the equation.

$2 \sqrt{b} + 5 + \sqrt{b} = 11$

Add $2 \sqrt{b}$ and $\sqrt{b}$ .

$3 \sqrt{b} + 5 = 11$

Move all terms not containing $\sqrt{b}$ to the right side of the equation.

Subtract $5$ from both sides of the equation.

$3 \sqrt{b} = 11 - 5$

Subtract $5$ from $11$ .

$3 \sqrt{b} = 6$

Divide each term in $3 \sqrt{b} = 6$ by $3$ and simplify.

Divide each term in $3 \sqrt{b} = 6$ by $3$ .

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

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $\sqrt{b}$ by $1$ .

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

Simplify the right side.

Divide $6$ by $3$ .

$\sqrt{b} = 2$

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

${\sqrt{b}}^{2} = 2^{2}$

Simplify each side of the equation.

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

${(b^{\frac{1}{2}})}^{2} = 2^{2}$

Simplify the left side.

Simplify ${(b^{\frac{1}{2}})}^{2}$ .

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

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

$b^{\frac{1}{2} \cdot 2} = 2^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$b^{\frac{1}{2} \cdot 2} = 2^{2}$

Rewrite the expression.

$b^{1} = 2^{2}$

Simplify.

$b = 2^{2}$

Simplify the right side.

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

$b = 4$