After expressing $k$ in terms of $x$, plug that expression into $x^2 + kx = 12$ so you have an equation with only one variable.

The quadratic you get will have two solutions. The question is not asking for each individual value, but for the sum of all possible $k$-values that come from those solutions.

For a quadratic equation $ax^2 + bx + c = 0$, the sum of its roots is $-b/a$. Use this idea to find the sum of the $x$-values, and then relate that to the sum of the $k$-values using $k = x + 4$.

In Desmos, let $y$ represent $k$. Enter the line as y = x + 4 (from $k - x = 4$) and the curve as x^2 + x*y = 12 (from $x^2 + ky = 12$ with $k$ replaced by $y$).

Use the intersection tool (tap the graph where they cross, or click the points that appear) to see the two intersection points. Note the $y$-coordinates of these points; these are the possible $k$-values.

In a new Desmos line, add the two $y$-values from the intersection points (for example, type y1 + y2 if you stored them, or directly enter their decimal values). The result of this addition is the sum of all possible values of $k$.

From the first equation

solve for $k$:

Now you can substitute this expression for $k$ into the second equation so everything is in terms of $x$ only.

Substitute $k = x + 4$ into $x^2 + kx = 12$:

Simplify the left side:

Divide the entire equation by $2$ to make the numbers smaller:

This quadratic gives the possible $x$-values.

For a quadratic of the form $ax^2 + bx + c = 0$, the sum of its roots is $-\dfrac{b}{a}$.

In $x^2 + 2x - 6 = 0$, we have $a = 1$ and $b = 2$, so the sum of the two possible $x$-values is

So, if the two $x$-values are $x_1$ and $x_2$, then $x_1 + x_2 = -2$.

Recall that $k = x + 4$.

If the two $x$-values are $x_1$ and $x_2$, then the corresponding $k$-values are

Add these:

From the previous step, $x_1 + x_2 = -2$, so

Therefore, the sum of all possible values of $k$ is $6$.