<PRE><B>Question 772194</B>
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Let *[tex \LARGE x] represent the side of the original square.  The rectangle formed by adding 5 to the length and 2 to the width must then have dimensions *[tex \LARGE (x\ +\ 5)] and *[tex \LARGE (x\ +\ 2)].  Since the area is the length times the width:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ +\ 5)(x\ +\ 2)\ =\ 130]


FOIL the binomials, collect like terms, put the quadratic in standard form, and then solve for *[tex \LARGE x].  Discard any negative root because a negative measure of length is absurd.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
&lt;font face=&quot;Math1&quot; size=&quot;+2&quot;&gt;Egw to Beta kai to Sigma&lt;/font&gt;
My calculator said it, I believe it, that settles it
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