Answer
$2$ different imaginary-number solutions that are complex conjugates.
Work Step by Step
$ x^{2}+4x+6=0\qquad$.... $a=1,\ b=4,\ c=6$
$ b^{2}-4ac\qquad$....substitute $b$ for $4,\ a$ for $1$ and $c$ for $6$
$=(4)^{2}-4\cdot 1\cdot 6$
$=16-24$
$=-8$
Since the discriminant is a negative number, there are $2$ different imaginary-number solutions that are complex conjugates.