Quadratic equation problem ax^2 + bx + c = 0 -- will give best answer.

Write a quadratic equation in the variable x having the given numbers as solutions.

Solutions are -10, 1

Working backward, if -10 & 1 are the solutions, then the factorization of the quadratic is

(x + 10)(x - 1) = 0

Notice if you put -10 or 1 in for x, you get 0!

So doing FOIL on this, we have

x*x + x*(-1) + 10*x + 10*(-1) = 0

x² - x + 10x - 10 = 0

x² + 9x - 10 = 0

can only be solved if either a,b or c is known
with only two points to pass through, an infinite number of parabolas exists that satisfies the above equation

the easiest way is assume a = 1, the simplest parabola

x^2 + bx + c = 0

substitute in solutions

100 -10b + c = 0
10b - c = 100

1 + b + c = 0
b + c = -1

add equations

11b = 99
b = 9

b + c = -1
c = -1 - b
= -1 - 9
c = -10

x^2 + 9x - 10

x=-10 therefore (x+10) is one factor
x=1 -------------(x-1) is another
multiply the factors (x+10)(x-1) you'll get x^+10x-x-10 which simplifies to x^+9x-10 which is the soln
x^+9x-10=0 is the reqd quad eqn

If those are the solutions then this must be true: (x + 10)(x - 1) = 0

Then use FOIL: x^2 + 9x - 10 = 0

x² + 9x - 10 = 0

(x+10)(x-1) = 0

x+10 = 0
x = -10

x-1 = 0
x = 1