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 $$k < 200/8.$$ 

I suppose you want the range of values of $k$ for which $kx^2+8x+k<6$ for all $\color{red}x$.
It is equivalent to $\;kx^2+8x+k-6<0$ for all $x$. Now a quadratic polynomial has a constant sign if and only if it has no real root, i.e. if and only if its (reduced) discriminant $\Delta'=16-k(k-6)<0$. Furthermore, this sign is the sign of the leading coefficient $k$. Thus, you have to solve the system of inequations:
$$ k^2-6k-16>0,\qquad k<0. $$
Can you continue?
