Values of $a$ for which $(a+4)x^2-2ax+2a-6 <0$ for all $x \in R$ How can we find all values of $a$ for which the inequality $(a+4)x^2-2ax+2a-6 <0$ is satisfied for all $x \in R$?
For the given condition, $D >0$, therefore $ (-2a)^2-4(2a-6)(a+4) >0$. Solving for $a$, I get $(a+6)(a-4) <0$, but the answer is $(-\infty, -6]$ which is not my answer.

Consider the families of quadratic functions $f_a(x)=(a+4)x^2-2ax+2a-6 $ we needs to find functions with their graph below $x$ axis these functions has maximum if $a+4<0$ and don't intersect the $x$ axis if discriminant is negative $D<0$ so we nedd to find solution of system of innequalities   
$$\begin{matrix}
a+4<0\\
D=4a^2-4(a+4)(2a-6)<0
\end{matrix}$$
$$\begin{matrix}a<-4 \\ 4a^2-8(a+4)(a-3)<0
\end{matrix}$$
$$\begin{matrix}a<-4 \\a^2-2a+24<0\end{matrix}$$
$$\begin{matrix}a<-4 \\a^2+2a-24>0\end{matrix}$$
$$\begin{matrix}a<-4\\a^2+6a-4a-24>0\end{matrix}$$
$$\begin{matrix}a<-4,\\(a-4)(a+6)>0\end{matrix}$$
$a+6<0\iff a\in(-\infty,-6)$ is solution
