algebra precalculus - Solving Radical Equations $x-7= sqrt{x-5}$

This the Pre-Calculus Problem:

$x-7= \sqrt{x-5}$

So far I did it like this and I'm not understanding If I did it wrong.

$(x-7)^2=\sqrt{x-5}^2$ - The Square root would cancel, leaving:

$(x-7)^2=x-5$ Then I F.O.I.L'ed the problem.

$(x-7)(x-7)=x-5$

$x^2-7x-7x+14=x-5$

$x^2-14x+14=x-5$

$x^2-14x-x+14=x-x-5$

$x^2-15x+14=-5$

$x^2-15x+14+5=-5+5$

$x^2-15x+19=0$

$(x-1)(x-19)=0$

Now this is where I'm stuck because when I tried to see if I got the right numbers in the parentheses I got this....

$x^2-19x-1x+19=0$

$x^2-20x+19=0$

As you may see I'm doing something bad because I don't get $x^2-15x+19$

Could anyone please help me and tell me what I'm doing wrong?