Three, (15 points) The three coefficients $a, b, c$ and the two roots $x_{1}, x_{2}$ of the quadratic equation $a x^{2}+b x+c$ are written in some order as five consecutive integers. Find all such quadratic equations, and briefly describe the reasoning process.
$$
\text { Three, } 2 x^{2}-2,-2 x^{2}+2 \text {. }
$$

According to the problem, by Vieta's formulas, we have
$$
\begin{array}{l}
b=-a\left(x_{1}+x_{2}\right), \\
c=a x_{1} x_{2} .
\end{array}
$$

Assume that among five consecutive integers, there is no 0, then these five consecutive integers must all be positive or all be negative, and their absolute values are five consecutive positive integers. Let $m$ be the smallest of these five absolute values, and $p$ be the largest.
Thus, $p-m=4$.
From equation (2), we get
$$
\begin{array}{l}
|c|=\left|a x_{1} x_{2}\right| \geqslant 1 \times 2 \times 3=6 \\
\Rightarrow m=p-4 \geqslant|c|-4 \geqslant 2 \\
\Rightarrow|c|>m^{3} .
\end{array}
$$

Then $4=p-m=|c|-m$
$$
\begin{array}{l}
>m^{3}-m=m\left(m^{2}-1\right) \\
\geqslant 2(4-1)=6>4,
\end{array}
$$

which is a contradiction.
Therefore, among the five consecutive integers, there must be 0.
If there is a root of 0, then from equation (2), we get $c=0$, which contradicts the five consecutive integers. Hence, both roots and $c$ are not 0. And $a$ is the coefficient of the quadratic term, so $a \neq 0$, this can only mean $b=0$.
Thus, from equation (1), we get
$$
x_{1}+x_{2}=0 \text {, and } x_{1}-x_{2} \leqslant p-m=4 \text {. }
$$

Therefore, there are two possible cases, i.e., the roots are $1, -1$ or $2, -2$.

In the first case, from equation (2), we have $a=-c$, and from the five consecutive integers, we get $a= \pm 2, c=\mp 2$.
Thus, the quadratic expressions are $2 x^{2}-2,-2 x^{2}+2$.
In the second case, from equation (2), we have $c=-4 a$, and the difference between $a$ and $c$ is greater than 4, which does not meet the requirements.