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

$9t^{2}+3t=0\qquad$.... $a=9,\ b=3,\ c=0$ $b^{2}-4ac\qquad$....substitute $b$ for $3,\ a$ for $9$ and $c$ for $0$ $=(3)^{2}-4\cdot 9\cdot(0)$ $=9+0$ $=9$ Since the discriminant is a positive number that is a perfect square, there are two rational solutions.