You can put this solution on YOUR website! (k+4)x^+(k+1)x+1=0 has equal roots.
in the quadratic equation ax^2 +bx +c the roots are equal when
the discriminant b^2-4ac = 0
comparing the equation
a= (k+4) b= (k+1) c=1
(k+1)^2-4(k+4)(1)=0
k^2+2k+1-4k-16=0
k^2-2k-15=0
k^2-5k+3k-15=0
k(k-5)+3(k-5)=0
(k-5)(k+3)=0
k- +5 OR k-3