In my previous question I asked what is $\sqrt{0}$ and learned that it is 0. I learned also that $\sqrt{2}^2=2$ in my class. Then I see $\sqrt{0}^2=0^2=0$. So in general to find $\sqrt{x}$ is it to find $y$ such that $y^2=x$?
Also i see it works for $\sqrt{4}=2$ because $\sqrt{4}^2=2^2=4$. So $\sqrt{4}$ is $y$ such that $y^2=4$.
It is true that $y = \sqrt{x}$ will satisfy $y^2 = x$. For real numbers, as long as $x \geq 0$ then $x$ has a real square root. The only thing you have to add to your explanation is that you actually find two solutions to $y^2 = x$ (except for $x=0$).
For example, as you say $\sqrt{2}^2 = 2$ but also $(-\sqrt{2})^2 = 2$. This is why you will often see that when people solve $y^2 = x$ they will write $y = \pm \sqrt{x}$, meaning that $y$ could be either of those values and satisfy the equation.
To summarize, we choose to write $\sqrt{x}$ as the positive solution $y$ to $y^2=x$, for $x >0$.