Answer
See explanations.
Work Step by Step
Step 1. Given $f(x)=2x^2-7x+4$, for $x=2$, we have $f(2)=2(2)^2-7(2)+4=-2\lt0$
Step 2. For $x=3$, we have $f(3)=2(3)^2-7(3)+4=1\gt0$
Step 3. As $f(2)$ and $f(3)$ have opposite signs, based on the intermediate value theorem, there is a real zero between 2 and 3.