Question 335131
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ =\ 6x\ +\ 27]


Put the equation in standard form


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 6x\ -\ 27\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3\ \times -9\ =\ -27]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3\ -\ 9\ =\ -6]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ +\ 3)(x\ -\ 9)\ =\ 0]


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


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 9]


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