Answer
There are $2$ different imaginary-number solutions that are complex conjugates.
Work Step by Step
$ 3a^{2}+5=-7a\qquad$....add $7a$ to each side to write in form of $ax^{2}+bx+c=0$.
$ 3a^{2}+7a+5=0\qquad$.... $a=3,\ b=7,\ c=5$
$ b^{2}-4ac\qquad$....substitute $b$ for $7,\ a$ for $3$ and $c$ for $5$
$=(7)^{2}-4\cdot 3\cdot(5)$
$=49-60$
$=-11$
Since the discriminant is a negative number, there are $2$ different imaginary-number solutions that are complex conjugates.