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?

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.
