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? 
 A: We can avoid squaring both sides. Let $x-5=u^2$, where $u \ge 0$.  Then $\sqrt{x-5}=u$. Also, $x=u^2+5$, so $x-7=u^2-2$. Thus our equation can be rewritten as
$$u^2-2=u, \quad\text{or equivalently}\quad u^2-u-2=0.$$
But
$$u^2-u-2=(u-2)(u+1).$$
Thus the solutions of $u^2-u-2=0$ are $u=2$ and $u=-1$.  Since $u \ge 0$, we reject the solution $u=-1$. 
We conclude that $u=2$, and therefore $x=u^2+5=9$.
A: $x-7= \sqrt{x-5}$
$(x-7)^2=\sqrt{x-5}^2$ 
$(x-7)^2=x-5$ 
$(x-7)(x-7)=x-5$
$x^2-7x-7x+49=x-5$
$x^2 - 15x + 54 = 0$
$(x - 9)(x - 6) = 0$
$x - 9 = 0 $ or $x - 6 = 0$
$x = 9$ or $x = 6$
Now check for extraneous solutions...
A: $x^2-14x+49=x-5$
$x^2-15x+54=0$
$x^2-9x-6x+54=0$
$x(x-9)-6(x-9)=0$
$(x-6)(x-9)=0$
$x=6$, or  $x=9$
A: $(x-7)=\sqrt{x-5}$
Check the domain first
$x-5 \geq0 \cap x-7 \geq0$
$$x \geq7$$
$(x-7)^2=\sqrt{x-5}^2$
$(x-7)^2=x-5$
$(x-7)(x-7)=x-5$
$x^2-7x-7x+49=x-5$
$x^2 - 15x + 54 = 0$
$(x - 9)(x - 6) = 0$
$x - 9 = 0 $ or $x - 6 = 0$
$x = 9$ or $x = 6$
checking the domain x=6 is an extraneous root.
x = 9
