Let f(x) = x^2-x-2

Start with the value 1. f(1) = 1^2 - 1 - 2 = -2 < 0

Next, we evaluate the value 2. f(2) = 2^2 - 2 -2 = 0 = 0

Hence we know that there exists a y in the interval [1,2] such that f(y) = 0.

Let's test more points, like 1.5

f(1.5) = (1.5)^2 - 1.5 - 2 = 2.25 - 1.5 - 2 < 0

Hence we know that there exists a y in the interval [1.5,2] such that f(y) = 0

You can continue this process iteratively to get your solution.

EDIT: omg, I made a terrible arithmetic error! sorry!