## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$2$ different imaginary-number solutions that are complex conjugates.
$x^{2}+7=0\qquad$.... $a=1,\ b=0,\ c=7$ $b^{2}-4ac\qquad$....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.