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 $0$?

From
$$\sqrt {x^2-3}=x-3$$
what we need stricktly as condition is that $x^2-3\ge 0$, then we can square both sides
$$\sqrt {x^2-3}=x-3\iff x^2-3=(x-3)^2$$
and for $x\neq 3$, noting that $x=3$ is not a solution, we obtain
$$\iff x^2-3=x^2-6x+9 \iff 6x=12\iff x=2$$
but since $x=2$ doesn't satisfy the equation it is to reject.
This is a legit way to solve, the extra solution come in from squaring both sides but we can exclude it at the end by checking directly on the original equation. 
As an alternative we can observe that since LHS is $\ge 0$ also RHS must be then $x-3\ge0 $ is required by the original equation.
