SOLUTION: The area of a rectangle is 65yd^2 and the length is 3yd less than double the width. What are tye dimensions

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Question 1119829: The area of a rectangle is 65yd^2 and the length is 3yd less than double the width. What are tye dimensions
Found 2 solutions by Alan3354, Shin123:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
The area of a rectangle is 65yd^2 and the length is 3yd less than double the width. What are tye dimensions
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Area = W*L = 65
W*(2W-3) = 65
2W^2 - 3W - 65 = 0
Solve for W, then for L

Answer by Shin123(626) About Me  (Show Source):
You can put this solution on YOUR website!
%282x-3%29%2Ax=65
2x%5E2-3x-65=0
%282x-13%29%28x%2B5%29=0
x=6.5 or x=-5 The dimensions are 6.5 and 10. Check: 6.5*2=13-3=10 6.5*10=65