Which values for b make the polynomial 16z2 + bz + 81 a perfect square trinomial?

As (a+b)^2 = a^2 + 2*a*b +b^2 (i)

here we have 
16z^2 + bz + 81 which can be written as (4z)^2 +  bz +9^2

from (i) we can see that in order to be perfect square bz should be
bz = 2* (4z) * (9)
bz= 72z

dividing z on both sides we get
b=72

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