#### Answer

$2$ different irrational solutions that are 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 positive number that is not a perfect square, there are $2$ different irrational solutions that are conjugates.