In the Wikipedia article on Field extensions, it says "$\mathbb{Q}(\sqrt 2) = \{a+b\sqrt 2 : a,b\in \mathbb Q\}$ is the smallest extension of $\mathbb Q$ containing all solutions to $x^2=2$."
Should this be $(bx-a)^2=2$ instead of $x^2=2$? I mean that this $(bx-a)^2=2$ consists of set $Q(\sqrt{2})$, which is smallest extension field of field $Q$?
No.
$\mathbb{Q}(\sqrt 2)$ is not dependent on any parameters.
You're missing the point of $a,b$ in the statement that $\mathbb{Q}(\sqrt 2) = \{a+b\sqrt 2 : a,b\in \mathbb Q\}$.
You need to have the real solutions of $x^2=2$ which are $\pm \sqrt 2$. Then you need to be closed under the field operations originally defined for $\mathbb Q$. This gives you arbitrary things $a+b\sqrt 2$ which should lie in the set, for any rational $a,b$.