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

We are interested in $f(x)=x^2+ax+a^2+6a<0\;\forall x \in (1,2)$
$$f(1)=1+a+a^2+6a \leq 0$$
That is $$a^2+7a+1\leq 0$$
$$ \frac{-7-\sqrt{45}}{2}\leq a \leq \frac{-7+\sqrt{45}}{2}$$
Also, we want, $$f(2)=4+2a+a^2+6a \leq 0$$
$$a^2+8a+4 \leq 0$$
$$\frac{-8-\sqrt{48}}{2} \leq a \leq \frac{-8+\sqrt{48}}{2}$$
Hence, overall, we need to take the intersection of the two constraint.$$ \frac{-7-\sqrt{45}}{2}\leq a \leq \frac{-8+\sqrt{48}}{2}$$
