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

There are $2$ different imaginary-number solutions that are complex conjugates.
$2a^{2}-3a=-5\qquad$....add $5$ to each side to write in form of $ax^{2}+bx+c=0$. $2a^{2}-3a+5=0\qquad$.... $a=2,\ b=-3,\ c=5$ $b^{2}-4ac\qquad$....substitute $b$ for $-3,\ a$ for $2$ and $c$ for $5$ $=(-3)^{2}-4\cdot 2\cdot(5)$ $=9-40$ $=-31$ Since the discriminant is a negative number, there are $2$ different imaginary-number solutions that are complex conjugates.