+152 I'm how to do this but this is my best shot. So first, you can think of it as a quadratic. So simplifying the expression you get \(x^2-kx-3x-k+6>0\)
Using the quadratic formula, you get \(\frac{-(-k+3) \sqrt {(-k-3)^2-4(6-k)}}{2} > 0\)
simplify, \(\frac{k-3\sqrt{k^2+6x+9-24+4k}}{2} = \frac{k-3\sqrt{k^2+10x-13}}{2} > 0\)
I hope that helped, i just kind of bashed it out, now i hope you can solve it. I dont really know how to myself. Sorry if this post was no help at all :(
-Evan 
+129852 This will be a parabola that turns upward
To find the x coordinate of the vertex we have that (k -3) / 2
So
[(k - 3) /2]^2 - (k -3) (k - 3)/2 - k + 6 > 0
( k^2 - 6k + 9 ] / 4 - ( k^2 - 6k+ 9)/2 - k + 6 > 0 multiply through by 4
k^2 - 6k + 9 - 2(k^2 - 6k + 9) - 4k + 24 > 0
k^2 - 6k + 9 - 2k^2 + 12k - 18 - 4k + 24 > 0
-k^2 + 2k + 15 > 0 mutiply through by -1 and reverse the inequality sign
k^2 - 2k - 15 < 0 factor
(k - 5) ( k + 3) < 0
We have three possible intervals that solve this
(-infinity, -3) or ( - 3, 5) or ( 5, infinity)
The outside intervals do not provide the correct solution
So.....the interval k = (-3, 5) is correct
