The general form of the quadratic equation is ax^2 + bx + c = 0.
You are being told, in effect, that 2 points on this curve are (-5,0) and (8,0).
Write out the quadratic equation three times, as follows:
a(-5)^2 + b(-5) + c = 0 and a(8)^2 + b(8) + c = 0. Both are = to 0 because y is 0 when x is either -5 or +8. Also, y = 90 = a(-2)^2 + b(-2) + c = 0, or
90 = 4a - 2b + c
Then we have 3 equations in 3 unknowns:
25a - 5b + c = 0
64a + 8b + c = 0
4a - 2b + c = 90
This is a system of linear equations, and there are various ways of solving such a system, including, but not limited to, substitution, matrices, determinants and the like. I solved this system on my TI-83 Plus calculator using matrix operations, and found that a = -3, b = -9 and c = 120.
Let me know if you need help in solving such a system.
Thus, the quadratic equation in question is
-3x^2 - 9x + 120 = y
Let's check this out! Does the point (8,0) satisfy -3x^2 - 9x + 120 = y ?
Does -3(8)^2 - 9(8) - 120 = 0 ? Does -3(64) - 72 + 120 = 0 ? Unfortunately, NO, which means that there is an arithmetic mistake somewhere in my presentation. Please read thru this work and ask any questions you may have; we will eventually solve the problem correctly.