ANSWER :
The zeros are 1/2 and -5
EXPLANATION :
From the problem, we have the function :
[tex]f(x)=2x^2+9x-5[/tex]
The zeros of the function are the values of x when f(x) = 0
[tex]2x^2+9x-5=0[/tex]
Using quadratic formula with a = 2, b = 9 and c = -5
[tex]\begin{gathered} x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ \\ x=\frac{-9\pm\sqrt{9^2-4(2)(-5)}}{2(2)} \\ \\ x=\frac{-9\pm\sqrt{81+40}}{4} \\ \\ x=\frac{-9\pm\sqrt{121}}{4} \\ \\ x=\frac{-9\pm11}{4} \\ \\ x=\frac{-9+11}{4}=\frac{1}{2} \\ \\ x=\frac{-9-11}{4}=-5 \end{gathered}[/tex]
