Answer:
h(x) = 2x^2 +14x -60
Explanation:
A quadratic is of the form
h(x) = ax^2 + bx +c
h(3) = h(-10) = 0
This tells us that the zeros are at x=3 and x = -10
We can write the equation in the form
h(x) = a( x-z1)(x-z2) where z1 and z2 are the zeros
h(x) = a(x-3) (x- -10)
h(x) = a(x-3) (x+10)
FOIL
h(x) = a( x^2 -3x+10x-30)
h(x) = a(x^2 +7x -30)
Let a = 2
h(x) = 2x^2 +14x -60