It asks you to consider where the parabola \(c^2 - 56c + 400\) is greater than zero. This will be at \(c < 28-8\sqrt{6}, c > 28 + 8\sqrt{6}\) (it's easier for you to understand and visualise if you draw this parabola out and check the inequality). Now, imagine that \(c = 8\), our function then becomes \(-2x+8\) which is a linear function which has exactly one intercept (and is not what we want in this case). Unfortunately, this solution lies in the subset of the reals \(c < 28-8\sqrt{6}\), so we omit \(c = 8\) only from that subset. Note that \(c < 8\) was already part of this subset of the real numbers, but we don't want to exclude it because it is only \(c = 8\) which is invalid.
If you like, you can think about it as \((c < 28-8\sqrt{6} \cup c > 28 + 8\sqrt{6}) \backslash \{c = 8\}\).