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.

The quadratic form$(a+4)x^2-2ax+2a-6=\begin{bmatrix} x & 1\end{bmatrix}\begin{bmatrix}a+4& -a \\ -a  & 2a-6\end{bmatrix}\begin{bmatrix}x\\ 1\end{bmatrix}$
So if you want the quadratic form to be negative for all $x$, you need a negative definite $\begin{bmatrix}a+4& -a \\ -a  & 2a-6\end{bmatrix}$. By Slvester's criterion, this implies $a^2+2a-24<0=(a+6)(a-4)<0$ and $a<-4$. Together these conditions mean $a\in(-\infty,-6)$.
