Question 68904
<pre>
What is asked in the problem?
Find the lenght and the width of the rectangle to the nearest thousandth.

Given:
Lenght of a rectangle is 3cm more than 2 times its width.
the area of the rectangle is 99 cm^2.

Representation:
Let     x = width of the rectangle
   2x + 3 = length of the rectangle

Write and equation: 
Remember that the Area of the rectangle = Length times the width

 A = (2x + 3)(x)
99 = 2x^2 + 3x
 0 = 2x^2 + 3x - 99

Factor : Use Quadratic Formula
    ax^2 + bx + c = 0, a = 2, b = 3, c= -99

       {{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
       {{{x = (-3 +- sqrt( 3^2-4*2*(99)))/(2*2) }}} 
       {{{x = (-3 +- sqrt( 9 + 792 ))/(4)}}} 
       {{{x = (-3 +- sqrt( 801))/(4)}}} 
       {{{x = (-3/4) +- (sqrt( 801)/4) }}}

x = 6.325   and     x = -7.825

no lenght or width that is negative so disregard x = -7.825

width of the rectangle = 6.325 cm   

lenght = 2x + 3 
       = 2(6.325) + 3 
       = 12.65 + 3
       = 15.65 cm