math question on equations + graphs

y=x² -2x-(k+3)
∵does not meet the x-axis
∴Δ<0
(-2)^2-4(1)[-(k+3)]<0
4-4(-k-3)<0
4-(-4k-12)<0
4+4k+12<0
4k<-16
k<-4
∴k<-4

the meaning of this question is the same as x^2 -2x-(k+3) = 0 does not have real root. that is the discriminant is < 0. The method of solving this problem can be refered to the answers above.

y=x² -2x-(k+3)
because there is no x-intersect
so Δ<0
(-2)² - 4(1)[-(k+3)]<0
4 + 4(k+3)<0
4(k+4)<0
k+4<0
k< -4

so the range of k is k< -4