1. ## imaginary solutions

3v^2 + 4v-1=0

2. ## Does this problem have a solution?

I believe that you are suppose to square the sides after adding the 1 to both sides, then solve...but it doesn't work out.

3v^2 + 4v-1=0
$v = \frac{-4\pm \sqrt{4^2 - 4(3)(-1)}}{2(3)}$
$v=\frac{-4\pm \sqrt{16 + 12}}{6}$
$\frac{-4+\sqrt{28}}{6},\frac{-4-\sqrt{28}}{6}$