The width of a rectangle is 3 less than twice its length. If the area of the rectangle is 131 cm^2, what is the length of the diagonal? The width of a rectangle is 3 less than twice its length. If the area of the rectangle is 131 cm^2, what is the length of the diagonal? 
I set up the basic equation to solve for l and I know I need to use Pythagorean therm to find the actual value but I keep getting the answer incorrect.
 A: width of rectangle = $w$
length of rectangle = $2w-3$
$w(2w-3) = 131$
$2w^2 - 3w - 131 = 0$
Solve for $w$ using the quadratic formula 
$w = -7.3779 $
$w = 8.8779$
We obviously only want the second answer. 
$w^2+l^2=d^2$
where d is the diagonal.
$\sqrt{w^2+l^2}=\sqrt{d^2}$
$d =\sqrt{w^2+(2w-3)^2}$
Plug in w to solve for d. 
A: If the width ($w$) of a rectangle is $3$ less than twice its length ($l$), then that translates into an equation: $w = 2l-3$. 
We also know that $w \cdot l = 131$. 
Then we're trying to find the length of the diagonal ($d$). Using the Pythagorean Theorem, that means we're trying to find $d=\sqrt{l^2 + w^2}$. 
Now we'll make some substitutions. We know that $w = 2l-3$ and $w\cdot l = 131$, so $2l-3 \cdot l = 131$. Do the multiplication: $2l^2 - 3l = 131$. That means $2l^2- 3l - 131=0$. You can solve this quadratic.
Now we rearrange: $w = 2l-3$ becomes $2l = w + 3$, so $l = \frac w 2 + \frac 3 2$. So $w \cdot \left(\frac w 2 + \frac 3 2\right) = 131$. Do the multiplication, then solve the quadratic.
Use the two results from solving the quadratics, and plug them into $d=\sqrt{l^2 + w^2}$. Then you're done.
(Note: when solving the quadratics, you'll get both positive and negative results. You'll want to use only the positive results.)
