(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
 A: If the solutions were supposed to be $-6$ and $1$, the problem would've been
$$
(x + 6)(x - 1) = 0\\
x^2 + 6x -x - 6 = 0\\
x^2 + 5x = 6
$$
Either you've copied the wrong problem, the wrong solution, or there is a mistake in your solution collection.
A: I am just going to do the thing first and we will talk about what you did after.
Note that $x^2-6x\sim (x-3)^2$. In fact
$$\begin{align}(x-3)^2=x^2-6x+9&=(x^2-6x)+9\\\Rightarrow x^2-6x&=(x-3)^2-9.\end{align}$$
Therefore if
$$x^2-6x=16$$ then
$$\begin{align}(x-3)^2-9&=16
\\\Rightarrow (x-3)^2&=25
\\ \Rightarrow x-3&=\pm5
\\ \Rightarrow x&=8\text{ or }-2.
\end{align}$$
The problems with what you did are as follows:


*

*If $x^2-6x=16$ it does not follow that $(x-3)^2=16$.

*You don't need to take the square roots of both sides necessarily. If $x^2=a$ then $x=\pm \sqrt{a}$ is a perfectly good implication.

*At the end you have two equations... $x-3=5$ and $x-3=-5$. To get rid of the three, remember you want $x=\dots$ add three to both sides, e.g.
$$\begin{align}
x-3&=5
\\\Rightarrow x-3+3&=5+3
\\\Rightarrow x+0&=8
\\ \Rightarrow x&=8.
\end{align}$$


You can do this because if two quantities are equal, and you add the same thing to both of them, then they are still equal.
A: $x^2 - 6x$ 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.
A: no ,no,what do you need is following
we have
$x^2-6*x=16$
$x^2-6*x+9=16+9$
$(x-3)^2=25$
now $x-3=5$ or  $x=8$
and $x-3=-5$  or $x=-2$
there is no mistake,why is answer $-6$?
$(-6)^2-6*(-6)=36+36=72$
