How to show this formula to get a square root of a number in "just few seconds" is true? I don't remember in which topic I found it but I know it was there. And I still have not find a proof of this nice approximation.

Let $x$ be a non perfect square number. If $y$ is the closer perfect square to $x$ such that $y < x$ then
$$\sqrt{x}\approx \sqrt{y}+\frac{x-y}{2\cdot \sqrt{y}}$$
And it gives at the maximum two correct decimals after the decimal point.

My first reaction to try to find from where this formula goes, was to expand it, I found $$2\sqrt{xy}\approx y +\sqrt{2y}\cdot(x-y)$$ and I tried to find a remarkable identity but I failed and I'm still stuck there. Also I don't know how should I prove the maximum of two correct digits after the comma.
If we look for example to the sqare root time of $1000$, $961$ is the closer perfect square which verifies the condition. Then we have $\sqrt{1000} \approx 31 + \frac{39}{2*31} = 31.62903...$ and with a calculator we have $\sqrt{1000} = 31.6227766017...$ which is quite good.
Any hints would be appreciate, thank you in advance.

\begin{align}
\sqrt{x} &= \sqrt{y + x - y}\\
&= \sqrt{y}{\left(1 + \frac{x-y}{y}\right)}^{1/2}\\
&\approx  \sqrt{y}\left(1 + \frac{x-y}{2 y}\right)\\
&= \sqrt{y} + \frac{x-y}{2 \sqrt{y}}
\end{align}
In going from the second to the third line above, I have used the Taylor series $\sqrt{1 + z} = 1 + \frac{z}{2} + O\left(z^2\right)$.
In fact this method is a good way of approximating square roots in mental arithmetic.
