Parabolas as conic sections

Still not quite sure if I am performing these right.

1. x - 4 = (y + 2)^2

x - 4 = y^2 + 4y + 4

x = y^2 + 4y + 8 (after this step, i am unsure if i am completing the square correctly)

does the step above equal to

x = (y^2 + 4y + 4) +8 + 4

x = (y + 2)^2 + 12

this would make the vertex (12, -2)? for some reason, this doesn't make sense. and then to find the "y" intercepts, i would just plug into the quadratic formula, correct?

a simpler one that is throwing me off is

x = 2y^2 + 3

do i complete the square with this one as well? I know that since you have to get this into a the form "x = a(y - k)^2 + h" the 2 signifies that this parabola will open up to the right. but as far as figuring out the vertex from only 2 terms, i am getting mixed up.

the last problem i have is the following

x = -y^2 + 4y - 3

since there is a negative sign in front of the y squared, do i go ahead and factor out a (-1), complete the square, and then plug into the quadratic formula? thanks again for everyone's help! hope you all had a wonderful weekend!

hapa