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User:LenBudney

I'm mostly experimenting with Wiki, but I'm very impressed with the quality of articles on this site!

You can visit my homepage at [1] (http://www.pobox.com/~lbudney/).

All Wikipedia text is available under the terms of the GNU Free Documentation License

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... b/(2a) from both sides, we get <math>x=\frac{-b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.</math> Generalizations The ...