Answer
The solutions are $-\displaystyle \frac{1}{2}+i\frac{\sqrt{19}}{2}$ and $-\displaystyle \frac{1}{2}-i\frac{\sqrt{19}}{2}$.
Work Step by Step
$ 0.4v^{2}+0.7v=0.3v-2\qquad$ ...write the expression in the form $ax^{2}+bx=c$ (add $-0.3v$ to each side)
$ 0.4v^{2}+0.4v=-2\qquad$ ...multiply the entire expression with $10$.
$ 4v^{2}+4v=-20\qquad$ ...divide each term with $4$.
$ v^{2}+v=-5\qquad$ ...square half the coefficient of $v$.
$(\displaystyle \frac{1}{2})^{2}=\frac{1}{4}\qquad$ ...complete the square by adding $\displaystyle \frac{1}{4}$ to each side of the expression
$ v^{2}+v+\displaystyle \frac{1}{4}=-5+\frac{1}{4}\qquad$ ...Write $v^{2}+v+\displaystyle \frac{1}{4}$ as a binomial squared.
$(v+\displaystyle \frac{1}{2})^{2}=-\frac{19}{4}\qquad$ ...take square roots of each side.
$ v+\displaystyle \frac{1}{2}=\pm\sqrt{-\frac{19}{4}}\qquad$ ...simplify $\displaystyle \sqrt{-\frac{19}{4}}=i\sqrt{\frac{19}{4}}=i\frac{\sqrt{19}}{2}$
$ v+\displaystyle \frac{1}{2}=\pm i\frac{\sqrt{19}}{2}\qquad$ ...add $-\displaystyle \frac{1}{2}$ to each side.
$v=-\displaystyle \frac{1}{2}\pm i\frac{\sqrt{19}}{2}$
