a.maths question

find the range of real values of k if the quadratic equation x^2+2kx+(6+k)=0 has
Δ=(2k)²-4(1)(6+k)
=4k²-24-4k
=4k²-4k-24

(a) real roots
Δ≧0
4k²-4k-24≧0
k²-k-6≧0
(k-3)(k+2)≧0
k≧3 or k≦ -2

(b) positive roots (unequal)
Δ＞0　and -2k＞０ and 6+k＞０
4k²-4k-24＞0 and k＜０ and k＞-6
k²-k-6＞0
(k-3)(k+2)＞0
[k＞3 or k＜ -2] and k＜０ and k＞-6
so no solution

(c) negative roots (unequal)
Δ＞0　and -2k＜０ and 6+k＞０
4k²-4k-24＞0 and k＞０ and k＞-6
k²-k-6＞0
(k-3)(k+2)＞0
[k＞3 or k＜ -2] and k＞０ and k＞-6
the solution is k＞３

(d) one positive root and negative roots
Δ＞0　 and 6+k＜０
4k²-4k-24＞0 and k＜-6
k²-k-6＞0
(k-3)(k+2)＞0
[k＞3 or k＜ -2] and k＜-6
the solution is k＜-6

(e) equal roots
Δ=0
4k²-4k-24=0
k²-k-6=0
(k-3)(k+2)=0
k=3 or k= -2

(f) equal positive roots
Δ=0 and -2k＞０and 6+k＞０
4k²-4k-24=0 and k＜０and k＞-6
k²-k-6=0
(k-3)(k+2)=0
[k=3 or k= -2 ]and k＜０and k＞-6
the solution is k= -2

(g) equal negative roots
Δ=0 and -2k＜０and 6+k＞０
4k²-4k-24=0 and k＞０and k＞-6
k²-k-6=0
(k-3)(k+2)=0
[k=3 or k= -2 ]and k＞０and k＞-6
the solution is k= 3

the range of real values of k if the quadratic equation x^2+2kx+(6+k)=0 has (d) one positive root and negative roots