After eliminating $y$, you will get an equation involving only $x$. Rearrange it into the standard quadratic form $ax^2 + bx + c = 0$.

For a quadratic $ax^2 + bx + c = 0$, you do not need to solve for each root separately to find their sum. Recall the relationship between $a$, $b$, and the sum of the roots $r_1 + r_2$.

In Desmos, enter the first equation as y = 10 - x^2 and the second equation as y = 3x - 2. These are equivalent to the original system and will appear as a parabola and a line.

Look for the points where the parabola and the line intersect. Click on each intersection point; Desmos will display their coordinates $(x_1, y_1)$ and $(x_2, y_2)$.

Note the two $x$-coordinates from the intersection points. In a new Desmos expression line, type their sum (for example, x1 + x2 using the numerical values shown) to see the total. That total is the sum of all possible values of $x$ that satisfy the system.

The system is

Because one equation has $+y$ and the other has $-y$, add the two equations so that $y$ cancels:

This simplifies to

Move 12 to the left side to put the equation in standard quadratic form:

This quadratic factors nicely. We need two numbers that multiply to $-12$ and add to $4$. Those numbers are $6$ and $-2$, so

Apply the zero-product property: $(x+6)(x-2)=0$ means $x+6=0$ or $x-2=0$, giving $x=-6$ or $x=2$.

The sum of all possible values of $x$ is:

So the correct answer is $-4$.