b²-4ac<0 so -4k²+36<0. But u dun have to find the value of X, just need to use discriminant to solve. So writing quadratic formula is not needed, only the part of b²-4ac is needed.

In this case, x² has negative coefficient so when equation<0 (bring RHS terms over to LHS to get LHS<0) it means the graph of this equation DOES NOT intersect x-axis, therefore b²-4ac<0 since there are no real solutions of x. As to why b²-4ac<0 is because the quadratic formula has a portion which includes √b²-4ac, and it is known that the √- number has no real solutions, therefore there are no real solutions of x if b²-4ac<0. Add Maths is very interesting and fun sometimes and is definitely a subject that I appreciate a lot.