Why is the solution to $x-\sqrt 4=0$ not $x=\pm 2$? If the equation is $x-\sqrt 4=0$, then $x=2$.
If the equation is $x^2-4=0$, then $x=\pm 2$.
Why is it not $x=\pm 2$ in the first equation?

The symbol $\sqrt{a}$ stands for the principal square root of $a$ which is always positive. Thus, the value of $x$ in the first equation is $2$.
The solution of the second equation, $x^2-4=0$, is given by
\begin{align*}
&\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\[2ex]
=\ &\frac{0\pm\sqrt{(0)^2-4(1)(-4)}}{2(1)}
\end{align*}
Simplyfing the above expression, we get  two answers: $+\sqrt{4}$
 and $-\sqrt{4}$. 
That's why the solution of the second equation is $\pm2$.
