Let's assume that roots of f(x) = 0 are x1, x2, x3. So we have to solve that x2 + 2x + 2 = xi. (i = 1, 2, 3). Since equation x2 + 2x + 2 - xi = 0 has not real roots, then its discriminant must be negative. That is Δ = 22 - 4(2 - xi ) < 0. Therefore xi < 1.
for example
if f(x) = x(x + 1)(x + 2), then a = 3, b = 2, c = 0.
if f(x) = x(x + 1)(x + 3), then a = 4, b = 3, c = 0.
We can see that a, b, c aren't constant. we don't find unique value of a,b,c.