How do you solve $-4(x+2)^2=-20$ ?

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Answer 1 · 2016-03-13T04:03:15

$x=-2+sqrt5$ and $x=-2-sqrt5$

From the given:
$-4(x+2)^2=-20$

divide both sides of the equation by $-4$

$(-4(x+2)^2)/(-4)=(-20)/(-4)$

$(cancel(-4)(x+2)^2)/cancel(-4)=(-20)/(-4)$

$(x+2)^2=5$

Extract the square root of both sides of the equation

$sqrt((x+2)^2)=+-sqrt5$

$(x+2)=+-sqrt5$

$x=-2+sqrt5$ and $x=-2-sqrt5$

God bless....I hope the explanation is useful.

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