After you set the right-hand sides equal, simplify to get a quadratic equation. Can you factor it, or use the relationship between a quadratic’s coefficients and the sum of its roots to find $x_1 + x_2$?

Once you know the $x$-coordinates, remember that for each point on the line $y = x - 2$, the $y$-value is 2 less than the $x$-value. How does that affect the sum of the two $y$-coordinates compared with the sum of the two $x$-coordinates?

In Desmos, enter y = x^2 - 7x + 10 on one line and y = x - 2 on another. You should see a parabola and a line that intersect at two points.

Click or tap on each intersection point that appears; Desmos will display the coordinates $(x_1, y_1)$ and $(x_2, y_2)$ of these points. Note the two $y$-values.

In a new expression line, type the sum of the two $y$-values you found (for example, y1 + y2 using the numeric values you read from the graph). The resulting value is the sum of the $y$-coordinates of the solutions.

At the intersection points, both equations give the same $y$-value, so set the right-hand sides equal:

Move everything to one side to get a quadratic in $x$:

The $x$-coordinates of the solutions are the roots of this quadratic.

Factor the quadratic:

So the $x$-coordinates of the intersection points are $x_1 = 2$ and $x_2 = 6$, and their sum is

(You could also use the fact that for $ax^2 + bx + c = 0$, the sum of the roots is $-\dfrac{b}{a}$, which here is $8$.)

Each solution must also satisfy the line equation $y = x - 2$, so if the $x$-coordinates are $x_1$ and $x_2$, the corresponding $y$-coordinates are

Add these to get the sum of the $y$-coordinates:

So, the sum of the $y$-coordinates of the solutions is $4$.