Question 188553
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Let <i>w</i> be the width.  Then the length is 2<i>w</i> + 6.  The area is the length times the width, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  w(2w + 6) = 56]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2w^2 + 6w - 56 = 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ w^2 + 3w - 28 = 0]


Factor the quadratic and exclude the extraneous negative root.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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