Same roots as coefficients in quadratic equation. 
Coefficients of quadratic equation $ax^2+bx+c=0$ are different
  integers. Roots of the equation are numbers $a$ and $b$. Find all
  equations that follow that condition.

After substitution $x_0=a,b$ we have two equations:
$$ a^3+ab+c=0 $$
$$ab^2+b^2+c=0$$
$$a^3+ab+c=ab^2+b^2+c$$
$$a^3+ab-ab^2-b^2=(a-b)(a^2+ab+b)=0$$
Solution $a=b$ doesn't count because of the assumption different integers. 
$$a^2+ab+b=0        $$
$$a^2=-b(a+1)       $$
The only conclusion I can think of is that one of $b$ or $a+1$ must be negative to prevent from contradiction $a^2<0$. Substition $a$ and $b$ into formula $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ leads to the same pair of equations.I don't know how to work this out.
Solution: $-2x^2+4x+16$

You already got
$$a^2+ab+b=0$$
from which we have
$$b=\frac{-a^2}{a+1}=\frac{(a+1)(-a+1)-1}{a+1}=-a+1-\frac{1}{a+1}$$
Since $\frac{1}{a+1}$ has to be an integer, we have to have $$a+1=\pm 1\implies a=0,-2$$
When $a=0$, we get $b=0=a$, so $a\not=0$. 
So, the answer is $(a,b)=(-2, 4)$ and $c=16$.
