Standard Equation of an Ellipse

I am having trouble finding the standard form of an ellipse, being given the following equation:

$\displaystyle

36x^2+9y^2+48x-36y+43=0

$

I know that I have to group terms and complete the square to make it fit the standard form of an ellipse:

$\displaystyle

\frac {(x-h)^2}{a^2} + \frac {(y-k)^2}{b^2} = 1

$

So I get to this and have been stuck:

$\displaystyle

(36x^2+48x+?)+9(y^2-4y+4)=-34+?+36

$

I don't think I can just factor out the 12, and have $\displaystyle

12(3x^2+4x+?) $, so I tried factoring out the 36:

$\displaystyle

36(x^2+\frac{4}{3}x+\frac{4}{9})

$

in which I got $\displaystyle

36(x+\frac{2}{83})^2 + 9(y-2)^2=18 $ <---which I got from $\displaystyle (-34+36(\frac{4}{9})+36)

$

So then I get

$\displaystyle

2(x+\frac{2}{3})^2 + \frac {(y-2)^2}{2} = 1

$

Does this mean that $\displaystyle a=\frac{1}{2}$ ? Is that even possible? And if so, does this mean that the center is $\displaystyle (-\frac{2}{3},2)$ ?

In this case then, the major axis is vertical, correct? So the foci would be $\displaystyle (-\frac{2}{3}, 2\pm\sqrt{1.5}) $ and the vertices would be $\displaystyle (-\frac{2}{3},2\pm\sqrt{\frac{1}{2}}) $ and $\displaystyle (-\frac{2}{3}\pm\sqrt{2},2)$ ?

Compared to what we've been learning in class (even though we haven't really even talked about it all that much), this all just seems too complicated to be correct, so I was didn't know if there was just something I was doing wrong or overthinking something.

Any help would be appreciated.