Applications of Differential Calculus (KUMON Level O) I don't understand what I need to do to solve this problem:

Show that equation $$\frac{x^2 - 3}{x-1}=0$$ has $2$ real solutions in $[-2,2]$.

I've found the derivative which is:
$$f'(x) = \frac{x^2 - 2x + 3}{(x-1)^2} > 0$$
The function is discontinuous at $x=1$. 
I don't know what to do next.
 A: Just solve 
$$\frac{x^2-3}{x-1}=0,$$
there is no need for calculus.
A: You have to solve $$\frac{x^2-3}{x-1}=0$$to solve it you can just multiply by $x-1$ both sides. $$x^2-3=0$$ hence the 2 solutions are $\pm\sqrt 3$. We know that there are no other solution because $\frac1{x-1}=0$ has no solutions and that $x^2-3$ is second degree thus it has at most $2$ roots
A: If you want to prove there are two roots in that interval, find two closed intervals$ [a,b] $where $f(a)f(b) <0$ and $f(x)$ is continuous. By Bolzano's Intermediate Value Theorem, there must be a root in each of those intervals
$[-2,-1] $ is one such interval since $f(-2)f(-1)= - \frac {1} {3} \cdot 1 \lt 0$
$[1.1,2]$ is another interval in which f has the same properties. 
Since both intervals are subsets of$ [-2,2]$, both have a root (by IVT), and their intersection is null, f must have two distinct roots in$ [-2,2]$.
Edit: must have AT LEAST two distinct roots. There could be more (not in this particular case, of course)
A: $Y’$ is alway greater than $0$ in the domain thus the function is always increasing in the domain at approaching $1$ from the negative side it is equal to infinity for the positive side it is equal to negative infinity and $f(-2)$ is smaller then $0$ so that is one real solution because the function is increasing  so it goes from some thing negative to infinity it will pass by $0$ and from negative infinity to posit if something it will go by $0$ too. thus there is two real solutions
