Question 330984
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First put your equation into Standard Form:


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


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


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ =\ 1\ \ ], *[tex \LARGE b\ =\ -2\ \ ], and *[tex \LARGE c\ =\ 5]



then recall the quadratic formula,


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


The discriminant, *[tex \Large \Delta], is equal to the portion of the quadratic formula under the radical, thus:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \Delta\ =\ b^2\ -\ 4ac]


Just plug in the numbers and do the arithmetic


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \Delta\ =\ (-2)^2\ -\ 4(1)(5)]


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