Answer
See explanations.
Work Step by Step
Step 1. Given $f(x)=2x^3-9x^2+x+20$, for $x=2$, we have $f(2)=2(2)^3-9(2)^2+(2)+20=2\gt0$
Step 2. For $x=2.5$, we have $f(2.5)=2(2.5)^3-9(2.5)^2+(2.5)+20=-2.5\lt0$
Step 3. As $f(2)$ and $f(2.5)$ have opposite signs, based on the intermediate value theorem, there is a real zero between 2 and 2.5.