<s>[INST] 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
 [/INST] 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.
</s>