Question 652996
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Put your equation into standard form, namely *[tex \LARGE ax^2\ +\ bx\ +\ c\ =\ 0]


Then calculate the discriminant.  *[tex \LARGE \Delta\ =\ b^2\ -\ 4ac]


If *[tex \LARGE \Delta\ <\ 0], then there are no real solutions.


If *[tex \LARGE \Delta\ =\ 0], there is one real solution with a multiplicity of 2, *[tex \LARGE x\ =\ \frac{-b}{2a}].


If *[tex \LARGE \Delta\ >\ 0], then there are two distinct real zeros. Use the quadratic formula:


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


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