Question 1094270
You made the incorrect substitution.  You found that u = 2, -1, 
but you substituted these values for x instead of u in the expression {{{ U =sqrt(x^2-3x) }}}
And we can reject u = -1 as a solution since a square root cannot be negative.
2 = {{{sqrt(x^2-3x) }}} -> x^2 - 3x - 4 = 0 -> (x-4)(x+1) = 0
So the two solutions are x = -1, x = 4