Question 328580
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For


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


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


For your problem:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ =\ 2\ \ ], *[tex \LARGE b\ =\ -5], and *[tex \LARGE c\ =\ -2]


so


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{-(-5)\ \pm\ \sqrt{(-5)^2\ -\ 4(2)(-2)}}{2(2)}]


Just do the arithmetic and simplify.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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