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
Best Answer
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.