√(3x+4)-√(2x-4)=2, squaring:
3x+4-2√[(3x+4)(2x-4)]+2x-4=4,
5x-2√[(3x+4)(2x-4)]=4,
5x-4=2√[(3x+4)(2x-4)], squaring again:
25x2-40x+16=4(6x2-4x-16)=24x2-16x-64,
x2-24x+80=0=(x-4)(x-20).
So x appears to be 4 or 20. We need to substitute these into the original equation:
x=4: √16-√4=4-2=2, so x=4 is a solution.
x=20: √64-√36=8-6=2, so x=20 is also a solution.