Why?

$(2x-3)^2 =9$.

Simple enough, I managed to deduce that $x=3$.

What I don't understand, however, is how $x$ can also equal $0$ in this instance, because if $2x-3=3$, and $x=0$, that's $2(0) -3$, which seems to equal $-3$ , rather than $3$

Could someone explain how $x$ could be $0$ in this instance?

Ahh yes. Of course. I always seem to overlook that. I must remember that the square root may also be the negative variant. Thank you.