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??