Quadratic Inequality in terms of variable $x$ 
Find the values of $a$ for which the inequality $x^2+ax+a^2+6a<0\;\forall x \in (1,2)$

$\bf{My\; Try::}$ We can Write Equation as $$x^2+ax+\frac{a^2}{4}+\frac{3a^2}{4}+6a<0$$
So $$\left(x+\frac{a}{2}\right)^2+\frac{3a^2+24a}{4}<0$$
Now how can i solve after that, Help required, Thanks

Rewrite the inequality as $$(2a+x+6)^2< (6-x)(3x+6)$$ 
The right hand side is positive for $1<x<2$, so we get
$$-\sqrt{(6-x)(3x+6)}<2a+x+6<\sqrt{(6-x)(3x+6)}$$
and
$$-x-6-\sqrt{(6-x)(3x+6)}<2a<-x-6+\sqrt{(6-x)(3x+6)}$$
$$\frac{-x-6-\sqrt{(6-x)(3x+6)}}{2}<a<\frac{-x-6+\sqrt{(6-x)(3x+6)}}{2}$$
The maximum of the right hand side is $-4+2 \sqrt{3}$ (let $x=2$) and the minimum of the left side is $-\frac{7}{2}-\frac{3 \sqrt{5}}{2}$ (let $x=1$)
so $$-\frac{7}{2}-\frac{3 \sqrt{5}}{2}<a<-4+2 \sqrt{3}$$
