Answer
There are $2$ different imaginary-number solutions that are complex conjugates.
Work Step by Step
$ 2a^{2}-3a=-5\qquad$....add $5$ to each side to write in form of $ax^{2}+bx+c=0$.
$ 2a^{2}-3a+5=0\qquad$.... $a=2,\ b=-3,\ c=5$
$ b^{2}-4ac\qquad$....substitute $b$ for $-3,\ a$ for $2$ and $c$ for $5$
$=(-3)^{2}-4\cdot 2\cdot(5)$
$=9-40$
$=-31$
Since the discriminant is a negative number, there are $2$ different imaginary-number solutions that are complex conjugates.