How do you convert 4y^2-3x-80y+424 into standard parabola equation?

Question

I know I did this wrong because the next step is to determine the vertex and I am way off for x.

Answer 1

Complete the square in y:
4y^2 -3x -80y +424 = 0
4y^2 -80y = 3x - 424
4(y^2 -20y + 100) = 3x - 424 +400
4(y - 10)^2 = 3x - 24
4(y - 10)^2 = 3(x - 8)
(y - 10)^2 = (3/4)(x - 8)
(y - 10)^2 = 4(3/16)(x - 8)
The vertex is at (8, 10)

Answer 2

4y^2 - 3x - 80y + 424 = 0
now take all y terms on one side and x on the other side
4y^2 - 80y = 3x - 424
dividing the equation by 4
y^2 - 20 y = 3/4 (x) - 106

now we try to make this equation in the form of (y-k)^2 = 4a (x-h)
y^2 - 2 (y) (10) = 3/4 (x) - 106

To make a complete square of y on the LHS we need to add 100 on both sides
y^2 - 2 (y) (10) + 100 = 3/4(x) - 106 + 100
(y - 10)^2 = 3/4 (x) - 6
(y-10)^2 = 3/4 ( x) - 3/4 (8) [ 6 = 3/4 (8)]
(y-10)^2 = 3/4 (x - 8)
(y - 10)^2 = 4 x (3/16) (x-8) [ equation in the form of (y-k)^2 = 4a(x-h)]

Now here (h,k) is the vertex of the parabola.
Therefore the vertex is (8, 10)