I read this as (k^2-k-20)/18.
What pair of factors of 20 differ by 1? This is essentially what the quadratic in k is asking. The pairs of factors of 20 are (1,20), (2,10), (4,5) and the only pair with a difference of 1 is (4,5). The larger of the two has a minus and the other a plus to make the difference -1 which is the coefficient of the middle term. So k^2-k-20 factorises: (k-5)(k+4). And we place this over 18: (k-5)(k+4)/18. This won't factorise any further.
When I was reviewing this question I felt there was more to it. Supposing the question was: what values can k take so that the expression is an integer? I spotted something interesting. The zeroes of the quadratic are 5 and -4: either of these values make the expression zero. But 5+4=9, and 9 is a factor of 18, the denominator. We can write the factors of the quadratic as y(y-9)/18, where y replaces k+4; or we can write y(y+9)/18, where y replaces k-5. The pairs of factors of 18 are (1,18), (2,9), (3,6). If the numerator contains any of these pairs, the expression will be an integer, or whole number, positive or negative.
What values of y must we have so that y(y-9)/18 is an integer? Let's start with y=0; the expression is zero, which is an integer. Now y=3; we have 3(-6)/18=-1, another integer. y=6; 6(-3)/18=-1. y=9; we get 0. y=12 we get 12*3/18=2. y=15; we get 15*6/18=5. In fact, all multiples of 3 work, so we can write y=3N as the general solution, where N is a positive or negative integer. What about k? We know that y=k+4 or k-5 so k+4=3N or k-5=3N are solutions. We can write these solutions as k=3N-4 or k=3N+5. If N=0 we have -4 and 5, which are the zeroes of the quadratic. If N=1, k=-1 or 8; if N=2, k=2 or 11; if N=3, k=5 or 14; if N=4, k=8 or 17; and so on.
If you substitute these and other values of k (according to the formula) you'll see that the expression is always an integer.