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

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

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)