Answer:
[tex]\large\boxed{-\dfrac{4}{3}\ and\ \dfrac{2}{3}}[/tex]
Step-by-step explanation:
[tex]f(x)=9x^2+6x-8\\\\\text{The zeros:}\\\\9x^2+6x-8=0\\\\9x^2+12x-6x-8=0\\\\3x(3x+4)-2(3x+4)=0\\\\(3x+4)(3x-2)=0\iff3x+4=0\ \vee\ 3x-2=0\\\\3x+4=0\qquad\text{subtract 4 from both sides}\\3x=-4\qquad\text{divide both sides by 3}\\\boxed{x=-\dfrac{4}{3}}\\\\3x-2=0\qquad\text{add 2 to both sides}\\3x=2\qquad\text{divide both sides by 3}\\\boxed{x=\dfrac{2}{3}}[/tex]
