The descriminant determines whether the zeroes are going to be complex or real. If the sign of the descriminant is negative, then complex zeroes will be found. In this case k^2-4k must be positive or zero for there to be real zeroes. So k^2-4k>=0. That means k(k-4)>=0, so k=0 or 4 satisfies this, but k cannot be zero, so k=4 is a solution. Now consider k(k-4)>0. This implies that k and k-4 are both negative or both positive: k<0 and k<4. So k<0 satisfies both inequalities because all negative numbers are less than 4. When k>0 and k>4 the inequality is satisfied, so we pick k>4 because 4 is greater than 0. So k<0 or k>=4 is the solution.
The solution is the union of the sets k<0 and k>=4, that is, all negative numbers and all positive numbers from 4 inclusive.
The image you have provided is too blurred to see properly, but I hope my answer helps.