Prove that $c<0$ and $4a+c<4b$
Since roots are complex $$4b^2-4ac<0$$ $$b^2< ac$$ Also $$2b>a+c$$ $$4b^2>a^2+c^2+2ac$$ $$b^2>\frac{(a^2+c^2)}{4}+\frac{ac}{2}$$ Then $$\frac{a^2+c^2}{4} < \frac{ac}{2}$$ $$(a-c)^2< 0$$ $$c> a$$
That’s all I could derive. How should I solve further
Let $f(x)= ax^2+2bx+c$ then $f(-1) = a-2b+c<0$, so $f$ is negative at $-1$
If $c>0$ then $f(0)= c>0$ so $f$ is positive at $0$ and since $f$ is continuous it must have real root. A contradiction so $c<0$
If $4a+c>4b$ then $f(-2) >0$ so $f$ is positive at $-2$ and $f$ has again real rooot. A contradiction again.