## College Algebra (10th Edition)

The solution set is {$\frac{1}{3}$}.
$9t^2-6t+1=0$ $a=9, b=-6, c=1$ $D=b^2-4ac=(-6)^2-4\cdot9\cdot1=36-36=0$. The equation has one real solution. $t=\frac{-b\pm\sqrt {b^2-4ac}}{2a}=\frac{6\pm\sqrt 0}{2\cdot9}$ Use the quadratic formula. $t=\frac{6}{18}$ $t=\frac{1}{3}$, The solution set is {$\frac{1}{3}$}.