Answer
$$v=2+\sqrt{13},\:v=2-\sqrt{13}$$
Work Step by Step
The quadratic formula states:
$$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$
Thus, solving the equation, we find:
$$ v_{1,\:2}=\frac{-\left(-4\right)\pm \sqrt{\left(-4\right)^2-4\cdot \:1\left(-9\right)}}{2\cdot \:1} \\ v=2+\sqrt{13},\:v=2-\sqrt{13}$$