SOLUTION: Hello! I am working on quadratic equations this week, in my Algebra class (section 11.4, #4). At the current moment I am stumped and totally confused on this one problem we are req

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Question 361286: Hello! I am working on quadratic equations this week, in my Algebra class (section 11.4, #4). At the current moment I am stumped and totally confused on this one problem we are required to do. Apart of the requirements for completing the assignment is to show every single step on how I got to my answer. I wasn't sure if anyone could help me, but I thought I would post to see who's the master at these type of equations. Thank you so much in advance for all your help!
Directions: Solve the equation by making an appropriate substitution.
4.)
2x-11√x- 40=0
the answer the book gives is Answer= {64}, but I don't see how I obtain the answer. Any help you can give is GREATLY appreciated.

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
2x+-+11sqrt%28x%29+-+40+=+0
You can sub y for sqrt(x)
2y%5E2+-+11y+-+40+=+0
(2y+5)*(y-8) = 0
y = 8
--> x = 64
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2y+5 = 0
y = -5/2
-->x = 25/4 This fits if you use the neg root.