Solve for n 0=18+128n^2

$0 = 18 + 128 n^{2}$

Rewrite the equation as $18 + 128 n^{2} = 0$ .

$18 + 128 n^{2} = 0$

Subtract $18$ from both sides of the equation.

$128 n^{2} = - 18$

Divide each term in $128 n^{2} = - 18$ by $128$ and simplify.

Divide each term in $128 n^{2} = - 18$ by $128$ .

$\frac{128 n^{2}}{128} = \frac{- 18}{128}$

Simplify the left side.

Cancel the common factor of $128$ .

Cancel the common factor.

$\frac{128 n^{2}}{128} = \frac{- 18}{128}$

Divide $n^{2}$ by $1$ .

$n^{2} = \frac{- 18}{128}$

Simplify the right side.

Cancel the common factor of $- 18$ and $128$ .

Factor $2$ out of $- 18$ .

$n^{2} = \frac{2 (- 9)}{128}$

Cancel the common factors.

Factor $2$ out of $128$ .

$n^{2} = \frac{2 \cdot - 9}{2 \cdot 64}$

Cancel the common factor.

$n^{2} = \frac{2 \cdot - 9}{2 \cdot 64}$

Rewrite the expression.

$n^{2} = \frac{- 9}{64}$

Move the negative in front of the fraction.

$n^{2} = - \frac{9}{64}$

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

$n = \pm \sqrt{- \frac{9}{64}}$

Simplify $\pm \sqrt{- \frac{9}{64}}$ .

Rewrite $- \frac{9}{64}$ as ${(\frac{3 i}{8})}^{2}$ .

$n = \pm \sqrt{{(\frac{3 i}{8})}^{2}}$

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

$n = \pm \frac{3 i}{8}$

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.

$n = \frac{3 i}{8}$

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

$n = - \frac{3 i}{8}$

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

$n = \frac{3 i}{8}, - \frac{3 i}{8}$