Find $a, b, c$ values from function $y=ax^2-bx+c$ and minimum value $D$ The problem reads like this:

The quadratic function which takes the value $41$ at $x = -2$ and $20$ at $x = 5$, is: $y = Ax^2-Bx+C$
The minimum value for this function is: $D$

I order it like this:
$$
\left\{
\begin{array}{1}
A(-2)^2-B(-2)+C=41\\
A(5)^2-B(5)+C=20
\end{array}
\right.
$$
$$
\left\{
\begin{array}{1}
4A+2B+C=41\\
25A-5B+C=20
\end{array}
\right.
$$
And try to solve it as a system of equations. But I'm stuck there. In fact, I'm not even sure if I'm approaching it the correct way.
EDIT
I got the answers for each variable, I just don't know the procedure to get them.
A=3; B=12; C=5; D=-7
 A: You need three equations to solve a linear system with three unknowns. The minimum equation will give you a third equation, assuming you are given the value of $D$ (or are allowed to express the other three vars in terms of $D$). Recall the vertex of a parabola occurs at $x=-b/2a$, or $x=B/2A$ in your case. Since there is a minimum, this must be it, as a parabola only has one turning point.
A: You can subtract the first equation from the second and get
$$
21A-7B=-21
$$
which gives $B=3A+3$ and so, substituting in the first equation,
$$
C=41-4A-2B=41-4A-6A-6=35-10A
$$
You are also given that the function has a minimum, so $A>0$. This minimum is taken at $x=B/(2A)$, so the minimum value is
$$
D=A\frac{B^2}{4A^2}-B\frac{B}{2A}+C=-\frac{B^2}{4A}+C=
-\frac{(3A+3)^2}{4A}+35-10A
$$
but the information is insufficient for determining $D$.
A: The question as given in the post does not make sense as the value can be as small as you want, according to this Desmos graph.
However, if we add in the restriction that $A, B, C$ are all positive integers, then we can solve the problem. Following on from egreg's solution, we need $A>0, B = 3A+3 ≥ 0$, which gives $A > -1$, and $C = 35-10A > 0$ which gives $A < 3.5$. Combining these two inequalities gives $-1 < A < 3.5$.
As $A$ increases, the value of $C$ decreases, which means the minimum value of the function decreases. Therefore, the maximum value of $A$ is $3$, which means $B=3(3)+3=12$, and $C = 35-10(3)=5$. The vertex thus has $x$-coordinate $\frac{-b}{2a} = \frac{3(3)+3}{2(3)} = 2$, so the minimum value is equal to $ax^2+bx+c = 3(2)^2- 12 \cdot (2) + 5 = -7$, exactly as described in your question.
