3*x^2 - 2*x + k = 0
f(x) = 3*x^2 - 2*x + k
As you recall, there exists a formula to find the roots of all quadratic equations in the form of a*x^2 + b*x + c = 0. The formula is,
x = (-b (+/-)(d)^(1/2)) / (2*a)
where d = b^2 - 4*a*c.
The only place in this formula that would give rise to an imaginary number (and hence, give us a solution with a non-zero imaginary component) is the radical, if and only if the quantity under it is negative. So, f(x) has roots with a non-zero imaginary component where d < 0 (d is negative). We are already given a and b. We'll plug those into the formula for d and substitute that into the inequality I stated above. Then, we'll solve for c, which is k in the original statement of the problem.
a = 3
b = -2
c = k
b^2 - 4*a*c < 0
4 - 4*3*k < 0
-12*k < -4
k > (1/3)
Therefore, f(x) has roots with a non-zero imaginary part where k > (1/3).
