No. This is not right.
Always remember that whenever you square both sides of a equation, you are incorporating some extra solutions.

In this case you got x < 21/6.
Check for x = 0, |x - 2| + |x - 3| = 2 + 3 = 5 > 2
Hence, your answer is wrong.

Probably the most easiest way to solve this problem is to visualize 2, 3, and x on the number line. Now, |x - 2| and |x - 3| are nothing but the distance of x from 2 and 3 on the number line. Hence, according to the given equation the sum of the distance of x from 2 and 3 on the number line must be less than 2.

From simple observation we can notice that if x lies between 2 and 3, the sum will be equal to 1, i.e. less than 2.

Now what about the points less than 2 or greater than 3?
Draw a number line as follows,

<--------------0----1----2----3----4------------>

Now, we can see that for any x less than 2 or greater than 3, the sum of the distances is (1 + twice the distance of x from 2 or 3 respectively).

Say, x is either d less than 2 or d greater than 3.
Then, the sum of distances = 1 + 2d < 2 ----> d < 0.5

So, (2 -0.5) < x < (3 + 0.5)
--> 1.5 < x < 3.5

This is the final solution.

Note: This approach may appear lengthy, but once someone grabs the basic idea behind it most of it can be done mentally and real fast.

shankar.ashwin wrote:|x-2| = x-2 (When x-2 is +ve) and
|x-2| = -x+2 (When x-2 is -ve)

...

x - 2 + x -3 < 2

I'm not sure if your understanding is wrong or you just typed it in short, but I'd suggest anyone following this method to be careful.

When (x - 2) is positive, then |x - 2| = (x - 2)
But that doesn't mean |x - 3| = (x - 3)

When x is greater than both 2 and 3, i.e. x > 3, then |x - 2| = (x - 2) and |x - 3| = (x - 3)

But when 2 < x < 3, then |x - 2| = (x - 2) and |x - 3| = -(x - 3) = (3 - x)

Luckily for this question such a logical mistake didn't create any problem.

However, for the other case, i.e. when (x - 2) is negative, i.e. x < 2, x is also less than 3 and (x - 3) is negative. Hence, |x - 2| = -(x - 2) and |x - 3| = -(x - 3)

The method followed by rijul007 is the proper algebraic way to tackle absolute value problems.