Suppose that a parabola and line are given by
$$y = 2x-k, y = x^2-(k+2)x+2k$$
and if one of the points at which they intersect is on $x$-axis, how can we find the ordinate of the other intersection point?
I approached the problem as follows:
$$x^2-(k+2)x+2k = 2x-k\iff x^2-(k+4)x+3k = 0$$
Whose roots are given by
$$x_{\pm} = \frac{(k+4)\pm \sqrt{(k+4)^2-12k}}{2}$$
If one of the intersection points lies on $x$-axis, there are two possibilities: $2x_{+}-k = 0$ or $2x_{-}-k = 0$. For the former condition,
$$4 + \sqrt{(k+4)^2-12k} = 0\implies 4 = -\underbrace{\sqrt{(k+4)^2-12k}}_{\geq 0}$$
Which is false, so $2x_{-}-k = 0$
$$4 = \sqrt{(k+4)^2-12k}$$
And which gives us $k = 0$ or $k = 4$. But we also have that
$$x^2_{-}-(k+2)x_{-}+2k = 0$$
You were nearly done. You found $k=0$ or $4$ and, the abscissa of the other intersection point being $$x_+=\frac{k+4+\sqrt{(k+4)^2-12k}}2,$$ its ordinate (in both cases $k=0,k=4$) is $$y_+=2x_+-k=4+\sqrt{(k+4)^2-12k}=8.$$