Answer
$$x=\frac{3+\sqrt{33}}{4},\:x=\frac{3-\sqrt{33}}{4}$$
Work Step by Step
Finding the zeros of the function using the quadratic formula, we find:
$$x_{1,\:2}=\frac{-\left(-3\right)\pm \sqrt{\left(-3\right)^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2} \\ x=\frac{3+\sqrt{33}}{4},\:x=\frac{3-\sqrt{33}}{4}$$
