(Highschool Pre-calculus) Solving quadratic via completing the square I'm trying to solve the following equation by completing the square:
$x^2 - 6x = 16$ 
The correct answer is -6,1. This is my attempt:
$x^2 - 6x = 16$
$(x - 3)^2 = 16$
$(x - 3)^2 = 25$ 
$\sqrt(x -3)^2 = \sqrt(25)$
$x - 3 = \pm5$
$x =\pm5 - 3$
$x = -8,2$
I did everything according to what I know,but my answer was obviously wrong. Any help is appreciated. Thanks

$x^2 - 6x$ isn't the same as $(x-3)^2$, so your equation $(x-3)^2 = 16$ is wrong. 
$(x-3)^2$ is actually $x^2 - 6x + 9$, so you should write 
$$x^2 - 6x + 9 = 25$$
and then 
$$(x-3)^2 = 25$$.
So your third equation is correct, even though your second wasn't.
I don't agree with your equation $\sqrt{(x-3)^2} = \sqrt{25}$. Technically it is correct, but you should know that in general $\sqrt{A^2}$ is not always $A$. In general, $\sqrt{A^2} = |A|$. So taking the square roots of both sides is not a good way to explain this. Instead, write $x-3 = \pm \sqrt{25}$. (It is a fact that if $z^2 = a$ and $a \geq 0$, then $z = \pm \sqrt{a}$.)
Your main mistake, and the only one that leads to an error in the result, is where you add 3 to the left side of your equation, but subtract 3 from the right side, and obtain $x = \pm 5 - 3$. You should instead add 3 to both sides. You need to do the same thing to both sides.
