Question 983831
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ ax^2\ +\ bx\ +\ c]


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{y\ -\ c}{a}\ =\ x^2\ +\ \frac{b}{a}x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{y\ -\ c}{a}\ +\ \frac{b^2}{4a^2}\ =\ x^2\ +\ \frac{b}{a}x\ +\ \frac{b^2}{4a^2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{y\ -\ c}{a}\ +\ \frac{b^2}{4a^2}\ =\ \left(x\ +\ \frac{b}{2a}\right)^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{b^2\ +\ 4a(y\ -\ c)}{4a^2}\ =\ \left(x\ +\ \frac{b}{2a}\right)^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ \frac{b}{2a}\ =\ \frac{\pm\ \sqrt{b^2\ +\ 4a(y\ -\ c)}}{2a}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{-b\ \pm\ \sqrt{b^2\ -\ 4a(c\ -\ y)}}{2a}]


Note that when *[tex \Large y\ =\ 0] this is just the Quadratic Formula.


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

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