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.