Question 986575
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Since you have used *[tex \Large a] as your independent variable in the original problem, I'm going to specify the standard form quadratic using Greek letter coefficients thereby avoiding confusion.


The standard form of a quadratic equation is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \alpha{x^2}\ +\ \beta{x}\ +\ \gamma\ =\ 0]


In your example, you are using *[tex \Large a] in place of *[tex \Large x]


And, comparing your example to the standard form, your *[tex \Large \alpha\ =\ 1], your *[tex \Large \beta\ =\ -6] and your *[tex \Large \gamma\ =\ 2]


The general solution of the standard quadratic equation is given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_1\ =\ \frac{-\beta\ +\ \sqrt{\beta^2\,-\,4\alpha\gamma}}{2\alpha}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_2\ =\ \frac{-\beta\ -\ \sqrt{\beta^2\,-\,4\alpha\gamma}}{2\alpha}]


Note:  All quadratics have two solutions in the Complex Numbers.


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

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