The length of a rectangle is 3 more than twice the width. the area of the rectangle is 119 square inches. what are the dimensions of the rectangle?

if x = the width of the rectangle, which of the following equations is used in the process of solving this problem?

A = L * Wx = W
L = 2x + 3
A = 119

119 = x(2x + 3)
119 = 2x^2 + 3x
2x^2 + 3x - 119 = 0
(2x + 17)(x - 7) = 0

2x + 17 = 0
2x = -17
x = - 17/2 or - 8.5wont work because it is negative

x - 7 = 0           L = 2x + 3
x = 7                L = 2(7) + 3
                        L = 14 + 3
                        L = 17

so the width (x) = 7 inches and the length (L) = 17 inches...
the equation used was : 119 = x(2x + 3)...or 2x^2 + 3x - 119 = 0...not sure which equation they want...but they are the same

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