use 3y^2-x-12y+14=0 to do the following

a. write the equation in standard form:

3y^2-x-12y+14=0

-x+14=-3y^2-12y

-x+14=-3(y^2+4y)

-x+14= -3[(y^2+4y+4)-4]

-x+14= -3[(y+2)^2-4]

-x+14=-3(y-2)^2+4

-x+10=-3(y-2)^2

-(x-10)= -3(y-2)^2/-1

x-10=3(x-2)^2

standard form:

x-10=3(x-2)^2

Is this right??