start making the function equal to 0,
[tex]12x^2-11x+2=0[/tex]use the quadratic formula in order to find the zeros of the function,
[tex]\begin{gathered} x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ a=12;b=-11;c=2 \\ x=\frac{-(-11)\pm\sqrt{(-11)^2-4(12)(2)}}{2(12)} \\ x=\frac{11\pm\sqrt{121-96}}{24} \\ x=\frac{11\pm\sqrt{25}}{24} \\ x_1=\frac{11+5}{24};x_2=\frac{11-5}{24} \\ x_1=\frac{2}{3};x_2=\frac{1}{4} \end{gathered}[/tex]finally, add both of the solutions
[tex]\frac{1}{4}+\frac{2}{3}=\frac{3+4\ast2}{12}=\frac{11}{12}[/tex]Answer:
The sum of the zeros for the function is 11/12
