Question 1061911
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ x\ -\ 2\ =\ 4]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ x\ -\ 6\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -3\ \times\ 2\ =\ -6]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -3\ +\ 2\ =\ -1]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ -\ 3)(x\ +\ 2)\ =\ 0]


So either


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ -\ 3\ =\ 0\ \Rightarrow\ x\ =\ 3]


or


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


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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