SOLUTION: If someone could help me with a step by step on how to solve this word problem in my quadratic equations chapter I would be most grateful! It says: The area of a rectangle in 48

Question 1041076: If someone could help me with a step by step on how to solve this word problem in my quadratic equations chapter I would be most grateful! It says:
The area of a rectangle in 48 in^2. It has w width that is 16 inches less than 4 times its length.
What is the length of the rectangle?
I think it wants me to set up a quadratic equation like:
4x^2-16-2304=0
but I'm not sure if I am on the right track? I have a few of these rectangle word problems that are really tripping me up!
Thanks in advance!
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
The area of a rectangle in 48 in^2. It has w width that is 16 inches less than 4 times its length.
What is the length of the rectangle?
-----
Length = L
Width = 4L-16
-------
Equation:
L(4L-16) = 48 in^2
------------------------
4L^2 - 16L - 48 = 0
------
L^2 - 4L - 12 = 0
---------
Factor::
(L-6)(L+2) = 0
-----
Positive solution::
Length = 6 inches
Width = 4L-16 = 24-16 = 8 inches
------------
Cheers,
Stan H.
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