Answer
$\{-3, -1, -\frac{1}{2}, 1 \}$
Work Step by Step
Step 1. Based on the Rational Zeros Theorem, list possible rational zeros $\frac{p}{q}=\pm1,\pm3,\pm\frac{1}{2},\pm\frac{3}{2}$
Step 2. Use synthetic division to find one or more zeros $x=1,-1$ as shown in the figure.
Step 3. Use to the quotient to solve $2x^2+7x+3=0$ or $(2x+1)(x+3)=0$, thus $x=-3, -\frac{1}{2}$
Step 4. List the real zero(s) $\{-3, -1, -\frac{1}{2}, 1 \}$