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

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

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Answer 1 · 2018-05-25T07:09:24

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

Explanation:

writing
$(2y+5)^2-5^2=0$
$(2y+5-5)(2y+5+5)=0$
this gives
$2y=0$
or
$2y+10=0$
this gives
$y=0$ or $y=-5$

Answer 2 · 2018-05-25T07:46:47

$y=0,-5$

Explanation:

Solve:

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

Take the square root of both sides.

$sqrt((2y+5)^2)=sqrt25$

$2y+5=+-5$

This gives us two equations:

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

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$color(blue)(2y+5=5$

Subtract $5$ from both sides.

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

$2y=0$

Divide both sides by $2$ .

$y=0/2$

$y=0$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

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

Subtract $5$ from both sides.

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

$2y=-5-5$

$2y=-10$

Divide both sides by $2$ .

$y=-10/2$

$y=-5$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Solutions for $y$ .

$y=0,-5$

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

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