Quadratic

Add 1 to both sides of the equation to make a quadratic: $(k+4)x^2 + (k+1)x+1=0$

The quadratic has exactly 1 solution when the discriminant($b^2 - 4ac$) equals 0. 

This means we have the equation: $(k+1)^2 - 4(k+4) = 0$

Simplifying the left-hand side gives: $k^2−2k−15=0$, which can be factored into $(k+3)(k−5)=0$, meaning $k = \color{brown}\boxed{-3, 5}$