Solve for I 15=2 square root of I-9

$15 = 2 \sqrt{I} - 9$

Rewrite the equation as $2 \sqrt{I} - 9 = 15$ .

$2 \sqrt{I} - 9 = 15$

Move all terms not containing $2 \sqrt{I}$ to the right side of the equation.

Add $9$ to both sides of the equation.

$2 \sqrt{I} = 15 + 9$

Add $15$ and $9$ .

$2 \sqrt{I} = 24$

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

${(2 \sqrt{I})}^{2} = 24^{2}$

Simplify each side of the equation.

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

${(2 I^{\frac{1}{2}})}^{2} = 24^{2}$

Simplify the left side.

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

Apply the product rule to $2 I^{\frac{1}{2}}$ .

$2^{2} {(I^{\frac{1}{2}})}^{2} = 24^{2}$

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

$4 {(I^{\frac{1}{2}})}^{2} = 24^{2}$

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

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

$4 I^{\frac{1}{2} \cdot 2} = 24^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$4 I^{\frac{1}{2} \cdot 2} = 24^{2}$

Rewrite the expression.

$4 I^{1} = 24^{2}$

Simplify.

$4 I = 24^{2}$

Simplify the right side.

Raise $24$ to the power of $2$ .

$4 I = 576$

Divide each term in $4 I = 576$ by $4$ and simplify.

Divide each term in $4 I = 576$ by $4$ .

$\frac{4 I}{4} = \frac{576}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 I}{4} = \frac{576}{4}$

Divide $I$ by $1$ .

$I = \frac{576}{4}$

Simplify the right side.

Divide $576$ by $4$ .

$I = 144$