SOLUTION: Solve by completing square (2y-3)^2=5

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Question 267507: Solve by completing square
(2y-3)^2=5

Answer by drk(1908) About Me  (Show Source):
You can put this solution on YOUR website!
step 1 - foil out the left side to get
(2y-3)(2y-3) = 5
first: 2y*2y = 4y^2
outer: 2y*(-3) = -6y
inner: (-3*2y = -6y
last: (-3)*(-3) = 9
combine the -6y and -6y to get -12y. So, we get
4y^2 - 12y + 9 = 5
step 2 - subtract 5 to get
4y^2 - 12y + 4 = 0
step 3 - divide by 4 to get
y^2 - 3y + 1 = 0
step 4 - take 1/2 of the middle term squared and add it o both sides to get
y^2 - 3y + (-3/2)^2 = -1 + (-3/2)^2
step 5 - writ the left side as a binomial squared and simplify the right side to get
(y-3/2)^2 = (5/4)
step 6 - take a square root to get
(y-3/2) = +-sqrt(5)/2
step 7 - add 3/2 to get
y = (3 +- sqrt(5))/2