Question 980756
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ \frac{2}{\sqrt{13}x}\ +\ \sqrt{13}\ =\ 0]


is not a quadratic, it is a cubic.  Are you sure you didn't mean


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ \frac{2}{\sqrt{13}}x\ +\ \sqrt{13}\ =\ 0]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 2\sqrt{13}x\ +\ \sqrt{13}]





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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \