Find the range of values of k for which x²+kx-3k<5 for some x, i.e. the curve y=x²+kx-3k goes below y=5

We know that x²+kx-3k-5<0 for some x for the values the k that we are trying to find.This will only occur when the curve has two distinct intersections with the x-axis. There are two distinct intersections only when the discriminant of the quadratic equation is more than 0. So k²-4(1)(-3k-5)>0 which is simplified as k²+12k+20>0.To find the values of k which satisfy this we solve k²+12k+20=0 by factorising and getting (k+10)(k+2)=0 so k=-10 or k=-2.Using these intersections with the x-axis we can sketch the graph of this quadratic and see that it is greater than 0 when k<-10 or k>-2. Therefore the solution to the question is k<-10 or k>-2.