This is easier than it looks. The square root term including x is only meaningful in this context when the quantity under the square root is strictly positive, so (x+3)(x-2)>0.
So we have just 2 conditions when this is positive: x+3 and x-2 are both positive; and x+3 and x-2 are both negative. Both positive means x+3>0 and x-2>0, so x>-3 and x>2. Since 2>-3, x>2 satisfies both inequalities.
Now when are they both negative? x+3<0 and x-2<0, so x<-3 and x<2. Since -3<2 then x<-3 satisfies both inequalities. So there you have it: M is real when x>2 or x<-3.
But if √2x is the numerator rather than √2 and the expression is √(2x) rather than (√2)x, we must eliminate x<-3 leaving x>2 as the solution.