Which is a zero of the quadratic function f(x) = 9x2 – 54x – 19?

F(x) = 9x² - 54x - 19 = 9x² - 57x + 3x - 19 = 3x(3x - 19) + 1(3x - 19)

= (3x - 19)(3x + 1)

f(x) = 0 if 3x - 19 =0 or 3x + 1 = 0

3x = 19 or 3x = -1

x = 19/3 or x = -1/3

(all credit goes to dalendrk)

Let's complete the square
isolate x terms
f(x)=(9x²-54x)-19
undistribute quadratic coefient
f(x)=9(x²-6x)-19
take 1/2 of linear coefient (-6) and square it
-6/2=-3, (-3)²=9
add positive and negative of it inside parenthasees
0=9(x²-6x+9-9)-19
copmlete the square
0=9((x-3)²-9)-19
distribute
0=9(x-3)²-81-19
0=9(x-3)²-100
add 100 to both sides
100=9(x-3)²
divide both sides by 9
100/9=(x-3)²
sqrt both sides, remember positive and negative roots
+/-10/3=x-3
rewrite 3 as 9/3
add 9/3 to both sides
9/3+/-10/3=x
so
9/3+10/3=x or 9/3-10/3=x
19/3=x or -1/3=x

the zeroes are at x=19/3 and -1/3

You have to give us options

What are the options?