Question 983243
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One of the numbers is *[tex \Large x]  The other one is *[tex \Large 2x\ +\ 2].  The product is *[tex \Large x(2x\ +\ 2)], which is to say *[tex \Large 2x^2\ +\ 2x].  Since the product is equal to 220,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x^2\ +\ 2x\ =\ 220].


Put the quadratic into standard form, factor it, and solve for *[tex \Large x] then calculate *[tex \Large 2x\ +\ 2]


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

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