Solve for y 9-4y^2=57

$9 - 4 y^{2} = 57$

Step 1

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

Subtract $9$ from both sides of the equation.

$- 4 y^{2} = 57 - 9$

Subtract $9$ from $57$ .

$- 4 y^{2} = 48$

Step 2

Divide each term in $- 4 y^{2} = 48$ by $- 4$ and simplify.

Divide each term in $- 4 y^{2} = 48$ by $- 4$ .

$\frac{- 4 y^{2}}{- 4} = \frac{48}{- 4}$

Simplify the left side.

Cancel the common factor of $- 4$ .

Cancel the common factor.

$\frac{- 4 y^{2}}{- 4} = \frac{48}{- 4}$

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

$y^{2} = \frac{48}{- 4}$

Simplify the right side.

Divide $48$ by $- 4$ .

$y^{2} = - 12$

Step 3

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

$y = \pm \sqrt{- 12}$

Step 4

Simplify $\pm \sqrt{- 12}$ .

Rewrite $- 12$ as $- 1 (12)$ .

$y = \pm \sqrt{- 1 (12)}$

Rewrite $\sqrt{- 1 (12)}$ as $\sqrt{- 1} \cdot \sqrt{12}$ .

$y = \pm \sqrt{- 1} \cdot \sqrt{12}$

Rewrite $\sqrt{- 1}$ as $i$ .

$y = \pm i \cdot \sqrt{12}$

Rewrite $12$ as $2^{2} \cdot 3$ .

Factor $4$ out of $12$ .

$y = \pm i \cdot \sqrt{4 (3)}$

Rewrite $4$ as $2^{2}$ .

$y = \pm i \cdot \sqrt{2^{2} \cdot 3}$

Pull terms out from under the radical.

$y = \pm i \cdot (2 \sqrt{3})$

Move $2$ to the left of $i$ .

$y = \pm 2 i \sqrt{3}$

Step 5

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.

$y = 2 i \sqrt{3}$

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

$y = - 2 i \sqrt{3}$

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

$y = 2 i \sqrt{3}, - 2 i \sqrt{3}$

Solve for y 9-4y^2=57

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