How to find f(x) - I suspect an error in the command

Hello

I want to solve a question, I don't understand why my answer is false so i think there might be an error in the order.

The quadratic function takes the value of 41 at x=-2 and the value of 20 at x=5. The function is minimalized at x=2.

I'm asked to give :

- the A,B,C of y=Ax^2+Bx+C

- the minimum value of this function, D

The vertex have coordinates (2;D)

2=-B/2A

B=-4A

I then replaced B in

41=A(-2)^2 +B(-2)+C

20=A(5)^2 +B(5)+C

In the end I've got f(x)= (-3/7)x^2-(12/7)x+(275/7)

It can't be right because A can't be an negative number if the function has a minimum. Also in this case the vertex is not in x=2 because D=239/7 (it's higher than f(-2)=20)

But when I calculate

f(-2)= (-3/7)(-2)^2-(12/7)(-2)+(275/7)

AND

f(5)=(-3/7)5^2-(12/7)5+(275/7)

It takes the value of 41 and 20.

Is it me confusing everything or there is really an error in the command?