How can you prove the uncertainty for position is:
$$\Delta{x} =\sqrt{\langle x^2\rangle-\langle x\rangle^2}$$
$\Delta{x}$, taken to be the root mean square of x.
$$\Delta{x} =\sqrt{\langle \left(x-\langle x\rangle\right)^2\rangle} $$
$$\Delta{x} =\sqrt{\langle \left(x-\langle x\rangle\right) \left(x-\langle{x}\rangle\right)\rangle}$$
$$\Delta{x} =\sqrt{\langle x^2-2x\langle x\rangle +\langle x \rangle^2\rangle}$$
This is the bit which I am not sure about and why I can do it (taking the outer braket and acting it on the inner x values:
$$\Delta{x} =\sqrt{\langle x^2\rangle -2\langle x \rangle \langle x\rangle +\langle x \rangle^2}$$
$$\Delta{x} =\sqrt{\langle x^2\rangle -2\langle x\rangle^2 +\langle x \rangle^2}$$
$$\Delta{x} =\sqrt{\langle x^2\rangle - \langle x \rangle^2}$$