What is a zero of the quadratic function f(x)=9x^2-54x-19

Short answer x = 19/3 and x=-1/3 <<<<< answer
The exact answer can be found with the quadratic formula. 

x = [tex] \frac{ -b\pm \sqrt{b^{2} - 4ac } }{2a} [/tex]

If you don't know about this procedure, this can only be solved by guess and by golly. I'll use that method

One thing is for certain: since 19 is prime the outside 2 numbers of the brackets can only be 1 and 19. So far what you have is this.

(x   19)(x   1) Now the problem is the signs and how the nine is divided up.
It has to get all the way up to 54 for the middle term. 9 times 19 is just too big so 3 and  3 is probably the way to do this.

(3x  19)(3x   1) We're getting closer. 3 * -19 is -57 so all we have to do is design something that takes 3x away. Or adds it.

(3x - 19)(3x  1) But what sign goes in the second factor?

So far what we have is
9x^2 - 57x +/- 1*3x  +/- 19 when we expand the two factors. So have have to fix the sign in the second factor. Try adding 1

(3x - 19)(3x + 1) I'm going to leave you to show that that works.

However that is not the answer. The answer is

3x - 19 = 0
3x = 19
x = 19/3

3x +1 = 0
3x = - 1
x = -1/3 and you are done.