Vivian and Ken are correct in that two of the roots are +/-3. However, you need to be careful "canceling" out a k on both sides, as this removes one of the possible roots. This type of problem is traditionally solved using factoring:
3k3 = 27k
Rearrange the equation so that it equals zero, just as you would do with a quadratic equation.
3k3 - 27k = 0
Factor out the LCM: 3k
3k(k2 - 9) = 0
Factor the quadratic expression in the parenthesis.
3k(k+3)(k-3) = 0
Set each part equal to 0 and solve for k.
3k = 0 --> k = 0 (this is the root that is removed when you "cancel" out a k early on)
k + 3 = 0 --> k = -3
k - 3 = 0 --> k = 3
This method gives you all 3 roots: k = -3, 0, and 3.