# Chapter 11 - Quadratic Functions and Equations - 11.3 Studying Solutions of Quadratic Equations - 11.3 Exercise Set - Page 718: 18

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.

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