Function Can Only be Solved by Simultaneous Equations, returns different/wrong answer each time it is solved? I'm having a lot of trouble with a specific question regarding functions, but I'm not sure where to post it.. the question is:

Let  $$ y = f(x) = a x^2 + bx + c $$ and have the values ($i \in \{1,2,3\}$): \begin{align} (x_i) &= (3,1,-2) \\ (y_i) &= (32, 6, -3) \end{align} what are $a$, $b$, $c$?

f(x) |-> ax^2 + bx + c, and {x: 3, 1, -2} and the range is, correspondingly, {y: 32, 6, -3} then what are the values of a, b and c?
So far I have done this by using simultaneous equations, but I get a different answer each time! For example, the first time I did it, I got $a = -13/10$, $b = 17/10$ and $c = 28/5$.
But if you plug these numbers in to the function, $f(1)\mapsto 6$, $f(-2) \mapsto -3$ but for some reason $f(3)$ does not return $32$?
The second time I did it, I got $a = 25/2$, $b= 31/6$ and $c = -35/3$ which literally did not work at all.
Help? Please?
(PS: I'm not sure if I'm allowed to post a question this simple here, since everyone else seems to be doing university-level math.. If I'm not, inform me and I'll take it down.)

My Work:
  4$a$-2$b$+$c$=-3
  -($a$+$b$+$c$=6)
  =3$a$-3$b$=9
9$a$+3$b$+$c$=32
  -(4$a$-2$b$+$c$=-3)
  =4$a$+4$b$=35
  =3$a$+3$b$=21 
3$a$-3$b$=9
  -(3$a$+3$b$=21)
  = -6$b$=-12 
therefore, $b$ = -2
4$a$-(-2 * -2) +c = (-3)
  4$a$+4+$c$=(-3)
  4$a$+$c$=-7
  -($a$-2+$c$=6)
  =3$a$=-11  
-2-(11/3)+$c$=6
  So.. $c$ = (35/3)? and $a$=-(11/3) and $b$ = -2?

Instead of writing my work I just did the problem again from scratch and lo and behold, a completely different answer set.. and this one's wrong, too.
 A: Hint:
In matrix form your system is
$$
y = X u 
$$
with the component-wise equations
$$
y_i = \sum_{j=1}^3 X_{ij} u_j
$$
with
$$
y = 
\left(
\begin{matrix}
y_1 \\
y_2 \\
y_3
\end{matrix}
\right)
\quad
X = 
\left(
\begin{matrix}
(x_1)^2 & x_1 & 1 \\
(x_2)^2 & x_2 & 1 \\
(x_3)^2 & x_3 & 1 \\
\end{matrix}
\right),
\quad
u = 
\left(
\begin{matrix}
a \\
b \\
c
\end{matrix}
\right)
$$
E.g.
$$
y_3 = (x_3)^2 a + x_3 b + 1 \cdot c
$$
which here is 
$$
-3 = (-2)^2 a + (-2) b + c = (-4,-2,1) \cdot (a,b,c)
$$
It has the solution
$$
u = X^{-1} y
$$
if the matrix $X$ is invertible. This is the case here, and the coefficients $a,b,c$ turn out to be integers.
A: we have to solve the system
$$a+b+c=6$$
$$4a-2b+c=-3$$
$$9a+3b+c=32$$
multiplying the first with $-4$ and adding to the second and multiplying the first by $-9$ and adding to the third we obtain
$$-6b-3c=-27$$
$$-6b-8c=-22$$
or
$$2b+c=9$$
$$3b+4c=1$$
multiplying the first by $-4$ and adding to the second we get 
$$-5b=-35$$ therefore we obtain $$b=7$$ can you proceed?
