Question 901578
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Use the quadratic formula:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{-b\ \pm\ \sqrt{b^2\ -\ 4ac}}{2a}]


letting 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ =\ a\ \ ], *[tex \LARGE b\ =\ -(2a\ +\ 1)\ \ ], and *[tex \LARGE c\ =\ a\ +\ 1]


All you have to do is simplify:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{-(-(2a\ +\ 1))\ \pm\ \sqrt{(-(2a\ +\ 1))^2\ -\ 4(a)(a\ +\ 1)}}{2a}]


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