Solve Using the Square Root Property (2y+5)^2=25

${(2 y + 5)}^{2} = 25$

Step 1

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

$2 y + 5 = \pm \sqrt{25}$

Step 2

Simplify $\pm \sqrt{25}$ .

Rewrite $25$ as $5^{2}$ .

$2 y + 5 = \pm \sqrt{5^{2}}$

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

$2 y + 5 = \pm 5$

Step 3

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.

$2 y + 5 = 5$

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

Subtract $5$ from both sides of the equation.

$2 y = 5 - 5$

Subtract $5$ from $5$ .

$2 y = 0$

Divide each term in $2 y = 0$ by $2$ and simplify.

Divide each term in $2 y = 0$ by $2$ .

$\frac{2 y}{2} = \frac{0}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 y}{2} = \frac{0}{2}$

Divide $y$ by $1$ .

$y = \frac{0}{2}$

Simplify the right side.

Divide $0$ by $2$ .

$y = 0$

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

$2 y + 5 = - 5$

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

Subtract $5$ from both sides of the equation.

$2 y = - 5 - 5$

Subtract $5$ from $- 5$ .

$2 y = - 10$

Divide each term in $2 y = - 10$ by $2$ and simplify.

Divide each term in $2 y = - 10$ by $2$ .

$\frac{2 y}{2} = \frac{- 10}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 y}{2} = \frac{- 10}{2}$

Divide $y$ by $1$ .

$y = \frac{- 10}{2}$

Simplify the right side.

Divide $- 10$ by $2$ .

$y = - 5$

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

$y = 0, - 5$

Solve Using the Square Root Property (2y+5)^2=25

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