Both $|x-3|$ and $\sqrt{2x-1}$ are nonnegative. What operation can you do to both sides to remove the absolute value bars and the square root symbol at the same time?

After you square both sides and simplify, you get a quadratic equation. Use factoring (if possible) or the quadratic formula, but remember that squaring can create extra solutions—so you must test each solution in the original equation.

In Desmos, enter y1 = abs(x-3) for the left-hand side and y2 = sqrt(2x-1) for the right-hand side. Remember that the square root is only defined for $x \ge 0.5$, so focus on that region of the graph.

Adjust the viewing window if needed so you can clearly see where the graphs of $y1$ and $y2$ cross. There should be two intersection points where the V-shaped graph of $|x-3|$ meets the curve $\sqrt{2x-1}$.

Tap each intersection point; Desmos will display its coordinates. The $x$-values of these intersection points are the solutions to $|x-3| = \sqrt{2x-1}$, and together they form the solution set.

Because of the square root, we must have $2x-1 \ge 0$, so $x \ge \tfrac12$.

Both sides of the equation

are nonnegative, so we can safely square both sides:

This gives

Expand and simplify:

Now we just need to solve the quadratic $x^2 - 8x + 10 = 0$.

For the quadratic $x^2 - 8x + 10 = 0$, the coefficients are $a=1$, $b=-8$, and $c=10$.

Use the quadratic formula

Substitute $a$, $b$, and $c$:

These are the two candidate solutions; next we simplify and check them in the original equation.

Simplify the expression from the quadratic formula:

So the candidates are $x = 4 - \sqrt{6}$ and $x = 4 + \sqrt{6}$.

Both are greater than $\tfrac12$, so they are in the domain of $\sqrt{2x-1}$. Check them in the original equation:

• For $x = 4 + \sqrt{6}$, $|x-3| = |1+\sqrt{6}| = 1+\sqrt{6}$ and

• For $x = 4 - \sqrt{6}$, $|x-3| = |1-\sqrt{6}| = \sqrt{6}-1$ and

In each case, left and right sides match, so both values are true solutions. Therefore, the solution set is $\{4 - \sqrt{6},\; 4 + \sqrt{6}\}$.