For what value of $k$, are the roots of the quadratic equation $$(k+4)x^2 + (k+1)x +1 = 0$$ equal.
Roots are equal, when the discriminant of this quadratic equation is zero.
$D = (k+1)^2-4(k+4)$
$0 = (k+1)^2-4(k+4) = k^2 - 2k - 15$
there are two solutions: $k = -3$ and $k = 5$