Question 618838
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 3x\ -\ 6]


Verify the lead coefficient is 1 then clear the constant to the RHS:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 3x\ \ \ \ =\ 6]


Divide the coefficient on the first degree term by 2, square the result, then add the result of the squaring operation to both sides.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 3x\ +\ \frac{9}{4}\ =\ 6\ +\ \frac{9}{4}]


Factor the perfect square in the LHS and do the arithmetic in the RHS:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(x\ -\ \frac{3}{2}\right)^2\ =\ \frac{33}{4}]


Take the square root of both sides.  Remember to consider both the postitive and negative roots.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ -\ \frac{3}{2}\ =\ \pm\frac{\sqrt{33}}{2}]


You should be able to clean it up from here.


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