SOLUTION: I am really having a hard time with these word problems and can use your help. The width of the rectangle is 13 ft less than twice its width. The area is 68 sq ft. Find length and

Question 998334: I am really having a hard time with these word problems and can use your help. The width of the rectangle is 13 ft less than twice its width. The area is 68 sq ft. Find length and width. I am trying to do this but I need assistance.
Found 2 solutions by fractalier, stanbon:
Answer by fractalier(6550) (Show Source): You can put this solution on YOUR website!
You will need to write your questions correctly. I cannot tell which is width and which is length.
But you can begin by stating
A = LW = 68
Then (I am guessing here)
L = 2W - 13

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
The width of the rectangle is 13 ft less than twice its width. The area is 68 sq ft. Find length and width.
Width:: w
Length:: 2w-13
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Equation:
width*length = 68 sq. ft
w(2w-13) = 68
2w^2 - 13w - 68 = 0
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Using the Quadratic Formula, width = 9.93 ft.
length = 2w-13 = 6.85 ft.
Cheers,
Stan H.
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