SOLUTION: This word problem involves quadratic equations which I am having a very difficult understanding. The width of a rectangle is 4 ft less than the length. The area is 12ft^2. Find the

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Question 167832: This word problem involves quadratic equations which I am having a very difficult understanding. The width of a rectangle is 4 ft less than the length. The area is 12ft^2. Find the length and width. I start out translating the equation like this:
A=LW
L=length
4-L=width
12ft^2=L(4-L) then I get lost from that point.

Answer by nerdybill(6948) About Me  (Show Source):
You can put this solution on YOUR website!
This word problem involves quadratic equations which I am having a very difficult understanding. The width of a rectangle is 4 ft less than the length. The area is 12ft^2. Find the length and width. I start out translating the equation like this:
A=LW
L=length
4-L=width <<--SHOULD BE L-4
.
Let x = length
then
x-4 = width
.
12 = x(x-4)
12 = x^2-4x
0 = x^2-4x-12
Factoring:
0 = (x-6)(x+2)
x = {-2, 6}
.
A negative solution doesn't make sense -- so, toss it out.
x = 6 feet (length)
.
Width:
x-4 = 6-4 = 2 feet (width)