Question 213061
The width of a rectangle is 9 less than twice its length. If the area of the rectangle is 100 centimeters squared, what is the length of the diagonal?
So far i have this:
W= 2L-9
L= ?
L(2L-9)=100
2L^2-9L-100=0 
.
Let L = length
then
2L-9 = width
.
L(2L-9) = 100
2L^2-9L = 100
2L^2-9L-100 = 0
.
You can solve it using the quadratic equation.  Doing so yields:
L = {9.67, -5.17}
We can throw out the negative solution leaving:
L = 9.67 centimeters
.
Width then is:
2L-9 = 2(9.67)-9 = 19.34-9 = 10.34 centimeters
.
To find the diagonal, you apply Pythagorean theorem:
19.34^2 + 9.67^2 = d^2
467.5445 = d^2
21.62 centimeters = d (diagonal)
.
Details of quadratic to follow:

*[invoke quadratic "L", 2, -9, -100 ]