Question 984419
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Let *[tex \Large w] represent the width.  Then *[tex \Large 2w\ +\ 3] represents the length.  Since the area of a rectangle is the length times the width,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2w^2\ +\ 3w\ =\ 160]


Which is to say


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2w^2\ +\ 3w\ -\ 160\ =\ 0]


Solve the quadratic for *[tex \Large w] and then calculate *[tex \Large 2w\ +\ 3]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \