How do you solve the equation #-4(x^2-8)=16#?

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Answer 1 · 2017-05-21T23:36:22

#x = +2 or x=-2#

There is no #x#=term, so we can solve this by finding the square roots.

#-4(x^2-8)=16" "larr div -4#

#x^2 -8= -4" "larr# add 8 to both sides

#x^2 = -4+8#

#x^2 = 4#

#x = +-sqrt4#

#x =+-2#