Question 1160251
<font face="Times New Roman" size="+2">


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  6x^2\ -\ 3y^2\ +\ 12x\ -\ 18y\ -\ 3\ =\ 0]


Move constant term to RHS, group by variable, and factor lead coefficient from each group:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  6(x^2\ +\ 2x\ \ \ \)\ -\ 3(y^2\ +\ 6y\ \ \ )\ =\ 3]


Complete the square on each group and compensate in the RHS considering the lead coefficient on each group:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  6(x^2\ +\ 2x\ +\ 1)\ -\ 3(y^2\ +\ 6y\ +\ 9)\ =\ 3\ +\ 6\ -\ 27\ =\ -18]


Factor the two perfect square trinomials, divide both sides by the constant term in the RHS, and arrange with positive term leading:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \frac{(y\,+\,3)^2}{6}\ -\ \frac{(x\,+\,1)^2}{3}\ =\ 1]

								
								
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 \ \ \ \ \ \ \ \ \ \  
								
{{n}\choose{r}}
</font>