Isolate the Root First

Squaring removes a root only when the root stands alone on its side. Try squaring $\sqrt{x+6}-1 = 3$ as it is. The left becomes ${\left(\sqrt{x+6}-1\right)}^{2}$, and a squared difference has a middle term: $x+6-2\sqrt{x+6}+1$. The root is still there, and the equation is messier than before.

So get the root alone first. Move the 1 to the other side: $\sqrt{x+6} = 4$. Now square both sides: $x+6 = 16$, so x is 10. Check in the original: $\sqrt{16}-1$ is 4 minus 1, which is 3.

The order is always the same: undo whatever is added to or subtracted from the root, square only when the root is alone, then check.