Answer
$2$ different imaginary-number solutions that are complex conjugates.
Work Step by Step
$ x^{2}-2x+4=0\qquad$.... $a=1,\ b=-2,\ c=4$
$ b^{2}-4ac\qquad$....substitute $b$ for $-2,\ a$ for $1$ and $c$ for $4$
$=(-2)^{2}-4\cdot 1\cdot 4$
$=4-16$
$=-12$
Since the discriminant is a negative number, there are $2$ different imaginary-number solutions that are complex conjugates.