Answer
$2$ different imaginary-number solutions that are complex conjugates.
Work Step by Step
$ x^{2}+7=0\qquad$.... $a=1,\ b=0,\ c=7$
$ b^{2}-4ac\qquad$....substitute $b$ for $0,\ a$ for $1$ and $c$ for $7$
$=(0)^{2}-4\cdot 1\cdot 7$
$=0-28$
$=-28$
Since the discriminant is a negative number, there are $2$ different imaginary-number solutions that are complex conjugates.