A rectangular parking lot has a width 30 feet more than its length. The owners are able to increase the width by 20 feet and the length by 40. The new lot has an area of 27,200 square feet, what is the area of the original lot?

The area of the original lot is 18000 sq. ft.

Let l be the length of the original rectangle.  Then l+30 is the width.  

The length of the second rectangle is increased by 40; this would be l+40.  The width of the second rectangle is increased by 20; this would be l+30+20=l+50.

The area of the second rectangle is found by multiplying its length and width; this gives us
(l+50)(l+40)=27200

Multiplying, we have
l*l+40*l+50*l+50*40=27200
l²+40l+50l+2000=27200
l²+90l+2000=27200

For quadratic equations, we want them equal to 0; subtract 27200 from both sides:
l²+90l-25200 = 0

We will use the quadratic formula to solve this:
[tex]l=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ \\=\frac{-90\pm \sqrt{90^2-4(1)(-25200)}}{2(1)} \\ \\=\frac{-90\pm \sqrt{8100--100800}}{2} \\ \\=\frac{-90\pm \sqrt{8100+100800}}{2}=\frac{-90\pm \sqrt{108900}}{2} \\ \\=\frac{-90\pm 330}{2}=\frac{-90-330}{2}\text{ or }\frac{-90+330}{2} \\ \\=\frac{-420}{2}\text{ or }\frac{240}{2}=-210\text{ or }120[/tex]

Since a negative length makes no sense, l=120.  This means the width is l+30=150, and the area is 120(150) = 18000.