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 $x^2 + kx = 12$, with $k$ replaced by $y$).

Use the intersection tool to identify the two points where the line and curve cross. The $y$-coordinates of these points are the possible values of $k$.

Add the two intersection-point $y$-coordinates in a new Desmos expression. This verifies the requested 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 $6$.