1. State the problem. We are given the quadratic function $y=2x^2-5x-3$ and a table where $p$ is the value when $x=-\tfrac{1}{2}$ and $q$ is the value when $x=2$. 2. Formula and rules. To find values from a quadratic evaluate by substitution using $y=2x^2-5x-3$. 3. Compute p. Evaluate at $x=-\tfrac{1}{2}$: $y(-\tfrac{1}{2})=2(-\tfrac{1}{2})^2-5(-\tfrac{1}{2})-3$. Simplify step by step: $=2(\tfrac{1}{4})+\tfrac{5}{2}-3$. Further simplification: $=\tfrac{1}{2}+\tfrac{5}{2}-3$. Combine terms: $=\tfrac{6}{2}-3$. Then $=3-3$. Therefore $p=0$. 4. Compute q. Evaluate at $x=2$: $y(2)=2(2)^2-5(2)-3$. Simplify: $=8-10-3$. Then $=-5$. Therefore $q=-5$. 5. Final answer. $p=0$, $q=-5$.