What are zeros of f(x)=x^2-3x-18. and explain so i can see if i can understand.

F(x) = x^2 - 3x - 18

Find the values that when multiplied equal -18 and when added equal -3 (6 and -3)
Plug it into f(x) = (x - 1st value)(x - 2nd value)

f(x) = (x - 6)(x + 3)
0 = (x - 6) (x + 3)

(One of the factors has to equal zero, so set them both equal to zero)

x - 6 = 0
x = 6

x + 3 = 0
x = -3

x = 6, x = -3

5/ 1/3=15 i think sorry if im wrong