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

$9t^{2}-48t+64=0\qquad$.... $a=9,\ b=-48,\ c=64$ $b^{2}-4ac\qquad$....substitute $b$ for $-48,\ a$ for $9$ and $c$ for $64$ $=(-48)^{2}-4\cdot 9\cdot 64$ $=2304-2304$ $=0$ Since the discriminant equals $0$, there is one repeated rational solution.
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