Use the first 3 points to set up 3 equations in 3 unknowns (a,b,c):
22 = a - b + c
8 = a + b + c
10 = 9a + 3b + c
Solve that for a, b, and c using the usual tricks: add the first two equations to get
30 = 2a + 2c, or a = 15 - c
Subtract the first two equations to get 14 = -2b, or b = -7.
Plug both of those into the first equation to get 22 = 9(15-c) - 3(-7) + c, after which you find c = 13. Then a = 2. Using those values, plug in the other two points, with missing x or y, and solve for what the p or q value must be to make the equation correct, now that you know what the coefficients a, b, and c are.
y=2x^2 - 7x + 13.
For the first unknown point, x = -2, so y = 2*(-2)^2-7(-2)+13 = 35.
For the second unknown point, you get
17 = 2q^2 -7q + 13. Using the quadratic equation,
q = {7 +/- sqrt(49-104)}/2*2. Unless you have a typo in the question, there is no real value q that satisfies the equation. There are two complex numbers that do, but I suspect you are not at the level of math yet where you are dealing with complex numbers. If you are, you know what to do from here.
