Squareroot of x equals -1
\sqrt{x}can only equal 0 or positive numbers (if we are looking for real solutions), so your problem has no solution.
Whenever you square both sides of an equation you can get erroneous roots. For example, suppose x=4 (so x equals 4 and no other number!!). If we square both sides we get {x}^{2}= 16. Solving for x, we get x= \pm 4. We must reject our erroneous solution of x = -4.
So, if we use the above step and square both sides we get x = 1. A quick checks yields that \sqrt{1}=-1, that is 1 = -1 which of course is absurd implying that there is no real solution
Right, so squaring both sides and bringing the 3 over to the RHS yields essentially the equation in the OP (y = x + 5 gets you there precisely, only in the variable y). I agree with the previous posters that there is no real solution. There are even issues with a complex solution. I think you'd have to define a different branch of the square root function in order for the equation to have a solution.
That's a complex analysis topic, though, so I don't think I'd worry about it for now, if you don't have to.