Question 649752
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This puppy won't factor over the rationals, so you are stuck with either completing the square or using the quadratic formula.  Since you said any method and the quadratic formula is easier, that's what you get:


The two roots of *[tex \LARGE ax^2\ +\ bx\ +\ c\ =\ 0] are given by 


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


For your problem (once you put it into standard form)


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ =\ 3\ \ ], *[tex \LARGE b\ =\ -8\ \ ], and *[tex \LARGE c\ =\ -4]


Plug in the numbers and simplify.


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