Solve for b square root of b+3=6

$\sqrt{b} + 3 = 6$

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

Subtract $3$ from both sides of the equation.

$\sqrt{b} = 6 - 3$

Subtract $3$ from $6$ .

$\sqrt{b} = 3$

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

${\sqrt{b}}^{2} = 3^{2}$

Simplify each side of the equation.

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

${(b^{\frac{1}{2}})}^{2} = 3^{2}$

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, ${(a^{m})}^{n} = a^{m n}$ .

$b^{\frac{1}{2} \cdot 2} = 3^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$b^{\frac{1}{2} \cdot 2} = 3^{2}$

Rewrite the expression.

$b^{1} = 3^{2}$

Simplify.

$b = 3^{2}$

Simplify the right side.

Raise $3$ to the power of $2$ .

$b = 9$