<s>[INST] 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 $0$?
 [/INST] When you solve an equation by squaring both sides, it's possible that extraneous solutions are introduced.  An extraneous solution is a value that you get at the end (like $x=2$ here) that doesn't actually solve the original equation.  This is exactly what happened here.  Since $x=2$ is the only value you got, and since it doesn't solve the original equation, that means the equation has no real solutions.
This can also be verified by graphing $y = \sqrt{x^2-3}$ and $y = x-3$ on the same axes and noticing that they never intersect.
Note also that $\left(\sqrt{x^2-3}\right)^2$ is just $x^2 - 3$, not $|x^2-3|$.  So considering $-(x^2 - 3) = (x-3)^2$ is technically incorrect, but you got lucky here in that it also has no real solutions.  Still, you should remove it completely from your work.
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