SOLUTION: Use completing the square to solve each equation. b^2 - 3b = 5 My answer is no where even close to what the book says it should be. I am in desperate need of help with these ty

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Question 44711This question is from textbook Intermediate Algebra
: Use completing the square to solve each equation.
b^2 - 3b = 5
My answer is no where even close to what the book says it should be. I am in desperate need of help with these types of problems. My book says the answer should be (3 + sqrt(29))/2 or (3 - sqrt(29))/2
This is what I came up with. What am I doing wrong?
b^2 - 3b = 5








This question is from textbook Intermediate Algebra

Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
This is what I came up with. What am I doing wrong?
b^2 - 3b = 5
You had b^2-3b+(3/2)^2= 5+(3/2)^2
You factored to get:
(b+(3/2))^2 = 5+(3/2)^2
You should have gotten:
(b-(3/2))^2 = 5+(3/2)^2
Cheers,
Stan H.






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