How do you solve #sqrt(4x+4)-8=0#?

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Answer 1 · 2016-01-30T07:34:58

#x=15#

#sqrt(4x-4)-8=0#

#rarrsqrt(4x-4)=8#

Square both sides

#rarr(sqrt(4x-4))^2=8^2#

#rarr4x-4=64#

#rarr4x=64-4#

#rarr4x=60#

#rarrx=60/4=15#