Question 39047
please help me solve:   3y^2-x-12y+14=0
write the equation in standard form

The standard form is (y-k)^2=4p(x-h)
1st, get the y's on the left and the rest on the right, as follows;
3y^2-12y=x-14
Complete the square on the left side and retain the equality , as follows:
3(y^2-4y+4)=x-14+3(4)
3(y-2)^2=x-2
Divide both sides by "3"  to get:
(y-2)^2=(1/3)(x-2)
Cheers,
Stan H.