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.
On
Let $f(x)=(a+4)x^2-2ax+2a-6$. Then $f(x)<0$ for all real $x$ precisely when $a<-4$ and $(2a)^2-4(a+4)(2a-6)<0$. The last inequality reduces to $(a+6)(a-4)>0$, so that $a<-6$ or $a>4$. Putting these together, we have the condition $a<-6$.
On
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
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 $a\in(-\infty,-6)$.