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 $0$?

$$\sqrt {x^2-3}=x-3$$
Since, by definition, $\sqrt {x^2-3} \ge 0$, we must have $x-3\ge 0$. That is we must have $x \ge 3$.
If $x \ge 3$, then $x^2-3 \ge 6$. So there will be no complications caused by squaring both sides of the equation.
\begin{align}
   x^2-3 &= x^2-6x + 9 \\
   6x &= 12 \\
   x &= 2
\end{align}
Since we need $x \ge 3$, there is no solution.
