Finding the average rate of change of $S(x) = -2x^2 + 14x - 12$ I have to show that the average rate of change of $S(x) = -2x^2 + 14x - 12$ in the interval $[x,x+h]$  is $-4x - 2h + 14$ and so far I did:
$$A(x) = \frac{S(x+h)-S(x)}{(x+h)-(x)} = \frac{(-2(x+h)^2+14(x+h)-12)-(-2x^2+14x-12)}{(x+h)-x} = \frac{-2(x+h)^2+14(x+h)-12+2x^2-14x+12}{-x+x+h} = \frac{-2(x+h)^2+14(x+h)+2x^2-14x}{h} = \frac{-2h^2+14h}{h} = \frac{-2h^2+14h}{h}\\ = -2^2 +14 = 4+ 14 = 18$$
I know this is totally wrong. What did I do wrong? How do I solve this?

Starting from
$$\frac{-2(x+h)^2+14(x+h)+2x^2-14x}h$$
expand all terms in the numerator and simplify:
$$=\frac{-2(x^2+2hx+h^2)+14x+14h+2x^2-14x}h$$
$$=\frac{-2x^2-4hx-2h^2+14h+2x^2}h$$
$$=\frac{-4hx-2h^2+14h}h$$
$$=-4x-2h+14$$
