Solution of system of equations involving $x_{1},x_{2},x_{3}$ 
Solve for $x_{1},x_{2},x_{3}$, given
$ax^2_{1}+bx_{1}+c=x_{2}$
$ax^2_{2}+bx_{2}+c=x_{3}$
$ax^2_{3}+bx_{3}+c=x_{1}$

Try: from $(1)$ and $(2)$
$a(x^2_{1}-x^2_{2})+b(x_{1}-x_{2})=(x_{2} - x_{3})$
And from $(2)$ and $(3)$
$a(x^2_{2}-x^2_{3})+b(x_{2}-x_{3})=(x_{3}-x_{1})$
Could some help me to find $x_{1},x_{2},x_{3}$, Thanks
 A: I would start with solving $$ax^2_{1}+bx_{1}+c=x_{2}$$
Via the quadratic formula,
$$x_1=\frac{-b\pm\sqrt{b^2-4a(c-x_2)}}{2a}$$
Repeating the same process, we have
$$x_2=\frac{-b\pm\sqrt{b^2-4a(c-x_3)}}{2a}$$
and
$$x_3=\frac{-b\pm\sqrt{b^2-4a(c-x_1)}}{2a}$$
Hence,
\begin{align}
x_1&=\frac{-b\pm\sqrt{b^2-4a(c-x_2)}}{2a}\\
&=\frac{-b\pm\sqrt{b^2-4a\left(c-\frac{-b\pm\sqrt{b^2-4a(c-x_3)}}{2a}\right)}}{2a}\\
&=\frac{-b\pm\sqrt{b^2-4a\left(c-\frac{-b\pm\sqrt{b^2-4a\left(c-\frac{-b\pm\sqrt{b^2-4a(c-x_1)}}{2a}\right)}}{2a}\right)}}{2a}\\
\end{align}
in which $x_1$ can be painfully solved for
to get
$$x_1=\frac{\pm\sqrt{-4 a c + b^2 - 2 b + 1} - b + 1}{2 a}$$
Then just repeat for $x_2$ and $x_3$ for which you will get
$$x_1=x_2=x_3=\frac{\pm\sqrt{-4 a c + b^2 - 2 b + 1} - b + 1}{2 a}$$
A: 
This is not a complete answer but the possible direction to solve this question:

$$a(x_i^2-x_j^2)+b(x_i-x_j)=x_j-x_k$$
$$\prod_{ijk} (x_i-x_j)[a(x_i+x_j)+b]=\prod_{ijk}(x_j-x_k)$$
$$\prod_{ijk} (x_i-x_j)=0 \quad \text{or} \quad \prod_{ijk}[a(x_j+x_k)+b]=1$$
where $i$, $j$ and $k$ have the convention of 
$\mathbf{e}_i\times \mathbf{e}_j=\mathbf{e}_k$.


*

*For the first condtion,
$$x_i=x_j \implies ax_i^2+bx_i+c=ax_j^2+bx_j+c \implies x_j=x_k$$
that is
$$x_i=x_j=x_k$$ which is easy to solve.

*Thanks to @Karn Watcharasupat's answer which reminds me the nested radical solutions,
$$
x=\frac{-b
\color{red}{\boldsymbol{\pm}} \sqrt{b^2-4ac-2b
\color{green}{\boldsymbol{\pm}} 2\sqrt{b^2-4ac-2b
\color{blue}{\boldsymbol{\pm}} 2\sqrt{b^2-4ac-2b
\color{red}{\boldsymbol{\pm}} \ldots}
}}}{2a}$$
The cyclic permutations of 
$\color{red}{\boldsymbol{+}}
 \color{green}{\boldsymbol{+}}
 \color{blue}{\boldsymbol{-}}
 \ldots$ or 
$\color{red}{\boldsymbol{-}}
 \color{green}{\boldsymbol{-}}
 \color{blue}{\boldsymbol{+}}
 \ldots$ gives distinct $x_i$, $x_j$, $x_k$.
The symmetric pattern of
$\color{red}{\boldsymbol{+}}
 \color{green}{\boldsymbol{+}}
 \color{blue}{\boldsymbol{+}}
 \ldots$ or 
$\color{red}{\boldsymbol{-}}
 \color{green}{\boldsymbol{-}}
 \color{blue}{\boldsymbol{-}}
 \ldots$ gives equal $x_i$, $x_j$, $x_k$.
See another question here.
A: option 1:
$x_1 = x_2 = x_3$
$ax^2 + bx + c = x\\
ax^2 + (b-1)x + c = 0\\
x= \frac {-(b-1) \pm \sqrt {(b-1)^2 - 4ac}}{2a}\\
$
option 2:
$x_1 \ne x_2 \ne x_3$
Find a polynomial that intersects:
$(x_1,x_2),(x_2,x_3),(x_3,x_1)\\
p(x) = x_2 + (x-x_1)\frac {x_3-x_2}{x_2-x_1} + (x-x_1)(x-x_2)\left(\frac {x_1 - x_2}{(x_3-x_2)(x_3-x_1)} - \frac {x_2-x_1}{(x_3-x_2)(x_2-x_1)}\right)$
