Set $x^2 - 4x$ equal to $2x - 4$ and rearrange all terms to one side to make a quadratic equation equal to $0$.

Once you have $x^2 - 6x + 4 = 0$, try to solve it. If it does not factor nicely with integers, use the quadratic formula.

You should get two possible $x$-values. Compare both of them to the answer choices and pick the one that appears.

In Desmos, enter y = x^2 - 4x as the first equation. You should see a parabola open upward.

Enter y = 2x - 4 as the second equation. This will draw a straight line on the same coordinate plane as the parabola.

Click on each point where the line and the parabola intersect. Desmos will show the coordinates of these intersection points; note the $x$-values and see which one matches one of the answer choices.

Because both equations equal $y$, any solution $(x, y)$ must make the right sides equal as well:

Now you have one equation in terms of $x$ only.

Move all terms to one side to get a quadratic equation equal to $0$:

So you need to solve $x^2 - 6x + 4 = 0$.

The quadratic formula for $ax^2 + bx + c = 0$ is

Here, $a = 1$, $b = -6$, and $c = 4$.

Substitute these values:

So the solutions for $x$ are of the form $x = \dfrac{6 \pm \sqrt{20}}{2}$.

First, simplify $\sqrt{20}$:

Now simplify the expression for $x$:

This means the two possible $x$-values are $3 + \sqrt{5}$ and $3 - \sqrt{5}$. Among the answer choices, only $3 + \sqrt{5}$ appears, so that is a value of $x$ that could be a solution to the system.