Chapter 11 - Quadratic Functions and Equations - 11.3 Studying Solutions of Quadratic Equations - 11.3 Exercise Set - Page 718: 16

$2$ different imaginary-number solutions that are complex conjugates.

$ 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.