Find the range of values of $k$ for which $kx^2 + 8x + k <6$ for all real values of $k$ 
Find the range of values of $k$ for which $kx^2 + 8x + k <6 $ for all real values of $k$.

I'm unsure if the discriminant must be greater than zero or less than zero.
My working steps: \begin{align}b^2 - 4ac = (8)^2 - 4(-2)(17-k) &> 0\\64 - 4(-2)(17-k) &> 0\\64 + 136 -8k  &> 0\\200 &> 8k\end{align} so my answer is $$k < 200/8.$$ 

We have
$$kx^2+8x+k<6 \iff kx^2+8x+k-6<0$$
and this is always true when $k<0$ and
$$b^2-4ac=64-4k(k-6)<0 \implies k^2-6k-16>0$$
that is $k<-2$.
