Bisection Method Numerical analysis Problem.

Question

I am trying to solve the following $$(\frac{x}{2})^2-\sin x = 0$$ with initial starting points $a_o = 1.5, b_0 = 2$ and $n = 1(1)5$ using Bisection Method.

From the little I have studied, I went about solving it like so;

Iteration 1

$f(x_1) = (\frac{1.5}{2})^2 - \sin (1.5)$ and this gave me $f(x_1) = 0.5$

Iteration 2

$f(x_2) = (\frac{2}{2})^2 - \sin (2)$ and this gave me $f(x_2) = 0.97$

The problem is, from what I read so far, either $f(x_1)$ or $f(x_2)$ has to be negative, so I concluded that I am doing it wrongly but I can't proceed.

Please, I need help on. Am not really a Mathematics student.