This question is solved applying the formula of the area of the rectangle, and finding it's width. To do this, we solve a quadratic equation, and we get that the cardboard has a width of 1.5 feet.

Area of a rectangle:

The area of rectangle of length l and width w is given by:

$A = wl$

w(2w + 3) = 9

From this, we get that:

$l = 2w + 3, A = 9$

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$ .

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$ , given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\Delta}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\Delta}}{2*a}$

$\Delta = b^{2} - 4ac$

In this question:

$w(2w+3) = 9$

$2w^2 + 3w - 9 = 0$

Thus a quadratic equation with $a = 2, b = 3, c = -9$

Then

$\Delta = 3^2 - 4(2)(-9) = 81$

$w_{1} = \frac{-3 + \sqrt{81}}{2*2} = 1.5$

$w_{2} = \frac{-3 - \sqrt{81}}{2*2} = -3$

Width is a positive measure, thus, the width of the cardboard is of 1.5 feet.

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