<PRE><B>Question 572838</B>
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{-b\ \pm\ \sqrt{b^2\ -\ 4ac}}{2a}]


for your problem:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ =\ 7\ \ ], *[tex \LARGE b\ =\ 9\ \ ], and *[tex \LARGE c\ =\ 4]


Plug in the numbers and do the arithmetic.  Since *[tex \LARGE 4ac\ &gt;\ b^2] you will end up with a conjugate pair of complex numbers.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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