SOLUTION: give exact and approximate solutions to three decimal places: y^2-8y+16=1

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Question 259994: give exact and approximate solutions to three decimal places: y^2-8y+16=1
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
y^2-8y+16=1
This = (y-4)^2= 1
Taking the square of both sides of the equation
we get y-4 = +1 or -1
therefore y =5 if it is +1 and y=3 if it is -1