Finding vertex by completing the square

Can anyone explain how to find the vertex of this equation by completing the square, please: y = - 3f^2 + 9f + 8.
I got as far as doing:
-3(f^2 - 3f) + 8
Then I'm not sure what to do now. Can anyone please show me how to get the answers and show full working? I know there's an easier method where you use the symmetry formula and sub it in but I need to do it by completing the square.

y = - 3f^2 + 9f + 8.
y/-3 = f^2 -3f - 8/3
y/-3 + 8/3 = f^2 - 3f
y/-3 + 8/3 + (-3/2)^2 = f^2 - 3f + (-3/2)^2
y/-3 + 8/3 + 9/4 = (f - 3/2)^2
y/-3 = (f - 3/2)^2 -8/3 - 9/4
y = -3(f - 3/2)^2 + 8 + 27/4
y = -3(f - 3/2)^2 + 59/4

Alternate forms:
3f(f-3)+y =8
3f^2+y=9f+8
......

y = - 3f^2 + 9f + 8.
=-3(f^2 - 3f) + 8
= -3{f^2 -3f +(-3/2)^2 - 9/4} +8 .
= -3{f-3/2}^2 +{8 +27/4}
= -3{f-3/2}^2 +{59/4}
The vetex is (3/2, 59/4} ................Ans

y = - 3f^2 + 9f + 8
y = - 3(f^2 - 3f) + 8
y = - 3(f^2 - 3f + 9/4) + 8 - 9/4
y = - 3(f - 3/2)^2 + 23/4

vertex is V(3/2,23/4) answer//