Confusion about solving quadratics $x^2-8x=5$ I got into a little confusion, hopefully someone can clear it up for me.
So I'm solving $x^2-8x=5$ right, and here are my steps:
$$1: x^2-8x=5$$
$$2: x^2-8x+16=21$$
$$3: (x-4)^2=21$$
$$4: x-4=\pm\sqrt{21}$$
$$x=\pm\sqrt{21} +4$$
So I have a question, why is it plus or minus? I thought the square root is usually positive?
 A: Every positive number $n$ has two square roots: $\sqrt{n}$ and $-\sqrt{n}$. Both $4 + \sqrt{21}$ and $4 - \sqrt{21}$ are solutions to this equation. You can try plugging them in with your calculator to verify.
What you're thinking of here is that $\sqrt{n}$ refers only to the positive square root. If you want the other square root, you must specify $-\sqrt{n}$. This way, we never need to be confused when we see $\sqrt{n}$, even though $n$ has two roots.
As for your other question, it is not true that $\sqrt{x} + y = \sqrt{x + y}$, which is what you're observing here.
A: There can be two square roots of every positive number: one positive and the other negative. For example, $ \sqrt 25 = ± 5$
 which is true as $ 5^2 = 25 $ and also, $ (-5)^2 = 25 $. 
Yes, the square root is usually positive ( $ \sqrt x $ denotes the positive root of $x$ only), but square roots can be negative as well. The above example proved it already. Also, in the fourth step you should write, 
$$ (x-4) = ± \sqrt 21 $$ 
A: 
First, why is it plus or minus?

Because the equation $\,x^2=b\,$ has two solutions $\,x = \,\pm \sqrt{b}\,$ when $\,b \gt 0\,$.

I thought the square root is usually positive?

That's correct. The real square root function $\,\sqrt{\,\cdot\,}\,$ returns the positive solution, by definition.

Second, and this really confuses me, we can see that in step 4 I removed the square from the factored side, what allows me to take the square root of the other side?

Suppose $\,a^2 = 21\,$, then that's equivalent to $\,a^2 - 21=0\,$, which using the algebraic identity $\,a^2-b^2=(a-b)(a+b)\,$ with $\,b=\sqrt{21}\,$ is in turn equivalent to $\,(a - \sqrt{21})(a+\sqrt{21})=0\,$. Therefore one of the factors must be $\,0\,$, so either $\,a=\sqrt{21}\,$ or $\,a=-\sqrt{21}\,$, which is usually written as $\,a = \pm \sqrt{21}\,$. Your question is about the case where $\,a = x-4\,$.
A: We have the equation
$$x^2 - 8x = 5$$
Add $16$ to both sides so we can factor the LHS in a nice way, so
\begin{align}
 x^2 - 8x + 16 &= 5 + 16 \\
 (x-4)^2 &= 21
\end{align}
Now take the square root on both sides (we can do this because both sides of the equation are positive terms)
$$ \sqrt{(x-4)^2} = \sqrt{21}$$ 
Notice that $x$ could be  a positive or negative number and still make the equation true,  we have to consider such cases, so
\begin{align}
x-4 &= \pm \sqrt{21} \\
x &= 4  \pm \sqrt{21}
\end{align}
will be the solutions to the quadratic equation.
