Question

(c-3)²=16 how do you solve this

Answer 1

c²-6c+9=16

c²-6c=16-9

c²-6c= 7= 0

c²-6c-7=0

c²-7c+c+7 =0

c(c-7) +1(c-7) =0

(c+1) or (c-7) = 0

c=-1 or c =7

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