Well, to take a sort of complex analysis point of view, you may want to consider the following:
$$\sqrt{1}=1$$$$\sqrt{1}=-1$$
Now where did we get this negative answer? We call it another branch. In fact, with algebra/precalculus, we usually stick to the primary branch where we have:
$$\sqrt{1}=1,\sqrt{1}\ne-1$$
This is simply used for less confusion to those who aren't well into complex analysis and similar things.
As for the solutions to your quadratic, I note that if we have the following:
$$x^2-5x+4=0$$
Then the solution is:
$$x=1,4$$
But, more interestingly, let me point at something more interesting: the quadratic formula:
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
Ever wonder why we have a $\pm\sqrt{}$ ? It is because, when dealing with something like a quadratic or any other polynomial, we are interested in ALL solutions, in particular, we are interested in both branches of the square root.
So plugging in $x=1,4$ may not appear to work for your original problem, but in a way, it does:
$$2-1=-\sqrt{1}$$
$$1=-\sqrt{1}$$
As I have noted all the way at the top of my answer, $\sqrt{1}=1,-1$ if we include all branches, such that we have:
$$1=-(-1)$$
This checks out.
However:
$$1\ne-(1)$$
Because that is simply the wrong branch. When we solved the quadratic using the quadratic formula, $x=1$ came when we used the square root as a negative. What this means is that to use $x=1$ as a solution, all square roots must come out negative.
For $x=4$, we must use positive square roots:
$$2-4=-\sqrt{4}$$
$$-2=-(2)$$
If we tried to use the wrong square root, we'd get the wrong answer:
$$-2\ne-(-2)$$
However, in a regular classroom environment or a class that does not involve high amounts of complex numbers or different roots, use only positive square roots because it is considered the primary branch.