Question 312667
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ ax^2\ +\ bx\ +\ c\ =\ 0]


ALWAYS has two solutions.  They might not be real numbers, but they are solutions nonetheless.


In this case,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 4x\ -\ 5\ =\ 0]


factors, so it has two rational roots.  Hint:  5 times -1 is -5 and 5 plus -1 = 4.



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