Question 596035
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You don't.  That is a whole set of quadratic equations.  The set has infinite elements; one for each possible real value of *[tex \LARGE y].  Without another relationship defined, you cannot determine which element of the set you are dealing with.


You can apply the quadratic formula to get an expression for the solution:


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


Where


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ =\ 1\ \ ], *[tex \LARGE b\ =\ -4\ \ ], and *[tex \LARGE c\ =\ -2y]


You will probably want to specify an interval for *[tex \Large y] such that the roots of the equation will have real values.


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