# How do you solve (2y + 5)^2 = 25?

May 25, 2018

$y = 0$ or $y = - 5$

#### Explanation:

writing
${\left(2 y + 5\right)}^{2} - {5}^{2} = 0$
$\left(2 y + 5 - 5\right) \left(2 y + 5 + 5\right) = 0$
this gives
$2 y = 0$
or
$2 y + 10 = 0$
this gives
$y = 0$ or $y = - 5$

May 25, 2018

$y = 0 , - 5$

#### Explanation:

Solve:

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

Take the square root of both sides.

$\sqrt{{\left(2 y + 5\right)}^{2}} = \sqrt{25}$

$2 y + 5 = \pm 5$

This gives us two equations:

color(blue)(2y+5=+5 and color(green)(2y+5=-5

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color(blue)(2y+5=5

Subtract $5$ from both sides.

$2 y + 5 - 5 = + 5 - 5$

$2 y = 0$

Divide both sides by $2$.

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

$y = 0$

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color(green)(2y+5=-5

Subtract $5$ from both sides.

$2 y + 5 - 5 = - 5 - 5$

$2 y = - 5 - 5$

$2 y = - 10$

Divide both sides by $2$.

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

$y = - 5$

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Solutions for $y$.

$y = 0 , - 5$