Solve $\sqrt {x^2-3}=x-3$ in $\mathbb R$ 
Solve $\sqrt {x^2-3}=x-3$ in $\mathbb R$

My attempt:
$|x^2-3|=(x-3)^2$
So $-(x^2-3)=(x-3)^2$ or $(x^2-3)=(x-3)^2$
If  $-(x^2-3)=(x-3)^2=x^2+9-6x$ 
So no solutions in $\mathbb R$
And if $(x^2-3)=(x-3)^2$
So $x^2-3=x^2+9-6x$
Now, can I delete $x^2$ with $x^2$ ? Like this 
$x^2-x^2-3-9+6x=0$
$6x=12$
$x=2$
But $f(2)$ isn’t equal to $0$?

I was taught to always find the domain of possible solutions first. We have 
\begin{cases}
x-3 \ge 0, \\
x^2-3 \ge0 
\end{cases} or \begin{cases}
x \ge 3, \\
-\sqrt{3} \le x \le \sqrt{3}
\end{cases}
which has no solutions 
