General form and standard form of a prabola

y^2 +  2y  + 12x + 25 = 0     

Subtract y^2 + 2y + 25 from both sides

12x = -y^2 - 2y -25        divide both sides by 12

x = (-1/12)y^2 - (1/6)y -  25/12      this is the general form

y^2 +  2y  + 12x + 25 = 0

Now....let's find the vertex form

Subtract   12x, 5 from both sides

y^2 + 2y   =    -12x  -  25

Complete the square on y

Take (1/2) of 2  = 1....square this  = 1^2  = 1 and add to both sides

y^2 + 2y + 1  =  -12x  - 25 + 1

y^2 + 2y + 1  =  -12x  - 24     factor both sides

(y + 1)^2  =  -12(x + 2)          

-12(x + 2) = (y + 1)^2

And this is in the vertex form      4p(x - h) = (y - k)^2