Since equation has two real roots $\alpha,\beta$
$\Rightarrow b^2-4ac > 0$
$\alpha < -2$ and $\beta > 2$ and $a > 0$
Product of roots < 0
$\Rightarrow \alpha\beta < 0\Rightarrow \large\frac{c}{a}$$ < 0$
$a > 0$
$\therefore C < 0$
$f(-1)=a-b+c < 0$
$f(1) =a+b+c < 0$
$\Rightarrow a+|b|+c < 0$
$f(2)=4a+2b+c < 0$
$f(-2)=4a-2b+c < 0$
$\Rightarrow 4a+2|b|+c < 0$
Hence (C) is the correct answer.