Answer
There are two rational solutions.
Work Step by Step
$ 10t^{2}-t-2=0\qquad$.... $a=10,\ b=-1,\ c=-2$
$ b^{2}-4ac\qquad$....substitute $b$ for $-1,\ a$ for $10$ and $c$ for $-2$
$=(-1)^{2}-4\cdot 10\cdot(-2)$
$=1+80$
$=81$
Since the discriminant is a positive number that is a perfect square, there are two rational solutions.