Okay, let's see. I need to solve the equation √(3x - 5) = 2. Hmm, square roots can sometimes be tricky, but I think the standard approach is to isolate the square root and then square both sides to eliminate it. Let me try that.

First, the equation is √(3x - 5) = 2. To get rid of the square root, I should square both sides. That should give me (√(3x - 5))² = 2². Simplifying both sides, that would be 3x - 5 = 4. Wait, is that right? Let me check. Squaring the left side removes the square root, leaving 3x - 5, and squaring the right side gives 4. So yes, that's correct.

Now, I need to solve for x. So starting with 3x - 5 = 4. Let me add 5 to both sides. That would give 3x = 4 + 5, which is 9. Then divide both sides by 3, so x = 9/3 = 3. Hmm, so x equals 3. But wait, I should check if this solution actually works in the original equation because sometimes squaring both sides can introduce extraneous solutions.

Let me plug x = 3 back into the original equation. The left side becomes √(3*3 - 5) = √(9 - 5) = √4 = 2. The right side is 2. So 2 = 2, which is true. Therefore, x = 3 is a valid solution. 

Wait, but the problem says "find all possible values of x." Since the square root function only outputs non-negative numbers, and the right side is 2, which is positive, there's only one solution here. So x = 3 is the only solution. 

Let me just recap the steps to make sure I didn't skip anything. Start with √(3x - 5) = 2. Square both sides: 3x - 5 = 4. Solve for x: x = 3. Check in original equation: works. So that's the answer. 

I think that covers everything. There's no need for additional steps or checking for other solutions because the square root function is one-to-one, so each input gives a unique output. Therefore, only one x satisfies the equation. 

Just to be thorough, let me consider if there's any other way this could be approached. For example, could there be a case where 3x - 5 is negative? But since we're taking the square root, the expression inside the square root must be non-negative. So 3x - 5 ≥ 0. Let's check with x = 3: 3*3 -5 = 4, which is positive. So that's okay. If there was a solution where 3x -5 was negative, it would be invalid, but in this case, it's not. 

Another way to think about it: the equation √(3x -5) = 2 implies that 3x -5 must equal 2 squared, which is 4. There's no other number that 3x -5 could be, because the square root function is injective (if a = b, then √a = √b and vice versa for non-negative a and b). So there's only one possible value for 3x -5, which is 4. Hence, only one solution for x. 

Therefore, I'm confident that x = 3 is the only solution.
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