Question 518536
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x^2\ +\ 2\ =\ 14x]


Put it into standard form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x^2\ -\ 14x\ +\ 2\ =\ 0]


Divide through by 2:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 7x\ +\ 1\ =\ 0]


Use the quadratic formula which is the general solution to any quadratic equation of the form:


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ =\ 1\ \ ], *[tex \LARGE b\ =\ -7\ \ ], and *[tex \LARGE c\ =\ 1]


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


Plug in the numbers and do the arithmetic.


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