Answer
There are $2$ different irrational solutions that are conjugates.
Work Step by Step
$ x^{2}+4x=8\qquad$....add $-8$ to each side to write in form of $ax^{2}+bx+c=0$.
$ x^{2}+4x-8=0\qquad$.... $a=1,\ b=4,\ c=-8$
$ b^{2}-4ac\qquad$....substitute $b$ for $4,\ a$ for $1$ and $c$ for $-8$
$=(4)^{2}-4\cdot 1\cdot(-8)$
$=16+32$
$=48$
Since the discriminant is a positive number that is not a perfect square, there are $2$ different irrational solutions that are conjugates.