SOLUTION: I am a little confused on how to solve this. I need to solve and give exact and approximate solution to three decimal places. Any help, showing steps, would be really appreciate

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Question 634449: I am a little confused on how to solve this. I need to solve and give exact and approximate solution to three decimal places. Any help, showing steps, would be really appreciated so I can understand how to do this type of problem. THanks.
y^2-18y+81=25
Found 2 solutions by ankor@dixie-net.com, josmiceli:
Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website!
y^2 - 18y + 81 = 25
subtract 25 from both sides and solve it as a quadratic equation
y^2 - 18y + 81 - 25 = 0
y^2 - 18y + 56 = 0
You can use the quadratic formula, but this will factor to
(y-4)(y-14) = 0
Two solutions which are integers
y = 4
and
y = 14

Answer by josmiceli(19441) (Show Source):
You can put this solution on YOUR website!

Complete the square

This ends up being what I started with,
since I didn't realize both sides were
perfect squares, but now I've proven it

Take the square root of both sides

and taking negative square root of

I can check by plugging both results back into equation

and

OK