Question 649588
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Use the quadratic formula:


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


Where


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ ax^2\ +\ bx\ +\ c\ =\ 0]


For your problem:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ =\ \ \ 1], *[tex \LARGE b\ =\ \ \ -1], and *[tex \LARGE c\ =\ 8]


Remember that *[tex \LARGE i^2\ =\ -1]


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