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.

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 $a<-6$.
