A certain function  contains points $(-3,5)$  and $(5,2)$. We are asked to find  this function,of course this will be simplest if we consider slope form equation 

$$y-y_1=m(x-x_1)$$

but could we find for general form of equation? for example quadratic? cubic?

Greg Martin and zyx have given you IMHO very good answers, but they rely on a few basic facts from Galois theory and/or group actions. Here is a more elementary but also a longer approach. 

Because we are in a field with $p$ elements, we know that $p$ is the characteristic of our field. Hence, the polynomial $g(x)=x^p-x$ has the property $$g(x_1+x_2)=g(x_1)+g(x_2)$$ whenever $x_1$ and $x_2$ are two elements of an extension field of $\mathbb{F}_p$. By little Fermat we know that $g(k)=k^p-k=0$ for all $k\in \Bbb{F}_p$. Therefore, if $r$ is one of the roots of $f(x)=x^p-x+a$, then $$f(r+k)=g(r+k)+a=g(r)+g(k)+a=f(r)+g(k)=0,$$ so all the elements $r+k$ with $k \in \Bbb{F}_p$ are roots of $f(x)$, and as there are $p$ of them, they must be all the roots. It sounds like you have already shown that $r$ cannot be an element of $\Bbb{F}_p$.

Now assume that $f(x)=f_1(x)f_2(x)$, where both factors $f_1(x),f_2(x)\in \Bbb{F}_p[x]$. From the above consideration we can deduce that
$$
f_1(x)=\prod_{k\in S}(x-(r+k)),
$$
where $S$ is some subset of the field $\Bbb{F}_p$. Write $\ell=|S|=\deg f_1(x)$. Expanding the product we see that
$$
f_1(x)=x^\ell-x^{\ell-1}\sum_{k\in S}(r+k)+\text{lower degree terms}.
$$
This polynomial was assumed to have coefficients in the field $\Bbb{F}_p$. From the above expansion we read that the coefficient of degree $\ell-1$ is $|S|\cdot r+\sum_{k\in S}k$. This is an element of $\Bbb{F}_p$, if and only if the term $|S|\cdot r\in\Bbb{F}_p$. Because $r\notin \Bbb{F}_p$, this can only happen if $|S|\cdot1_{\Bbb{F}_p}=0_{\Bbb{F}_p}$. In other words $f_1(x)$ must be either of degree zero or of degree $p$.
