Answer
See explanations.
Work Step by Step
Step 1. Given $f(x)=2x^3-5x^2-5x+7$, for $x=0$, we have $f(0)=2(0)^3-5(0)^2-5(0)+7=7\gt0$
Step 2. For $x=1$, we have $f(1)=2(1)^3-5(1)^2-5(1)+7=-1\lt0$
Step 3. As $f(0)$ and $f(1)$ have opposite signs, based on the intermediate value theorem, there is a real zero between 1 and 2.