I need help with a math pre-calc question(find vertices,foci & eccentricity) full points for the best answer!

Find the center, vertices, foci and eccentricity of the ellipse and sketch its graph. 9x^2+4y^2+36x-24y+36=0

Please help me :) Thank you!

Regroup x terms and y terms
9x^2+36x+4y^2+24y=-36
factorout the lead coefficient
9(x^2+ 4x+___)+4(y^2+6y+___)=-36+____+_____
now complete the squares
Half of the "b" then square it. But distribute to know what to add on the right side
(1/2)(4)=2, 2^2=4, so add 4 inside the parens, but 9(4)=36, add this on the right
9(x^2+4x+4)+4(y^2+6y+9)=-36+36+36
simplify and factor
9(x+2)^2+4(y+3)^2=36
you must have 1 on the right, so divide by 36
[(x+2)^2]/4+[(y+3)^2]/9=1
So the center is at (-2,-3)
a=sqr(9)=3
b=sqr(4)=2
a^2-b^2=c^2
so c= sqr(5)
Since the large denominator is under y, the ellipse is vertical
So the major vertices are (-2,-6) and (-2,0)
The minor vertices are (-4,-3) and(0,-3)
The foci are on the major axis so they are(-2,-3+/-sqr(5))
Eccentricity=e=c/a=sqr(5)/3