Approximate My Squares Inspired by this video by tecmath.

An approximation of the square root of any number x can be found by taking the integer square root s (i.e. the largest integer such that s * s ≤ x) and then calculating s + (x - s^2) / (2 * s). Let us call this approximation S(x). (Note: this is equivalent to applying one step of the Newton-Raphson method).

Though this does have quirk, where S(n^2 - 1) will always be √(n^2), but generally it will be very accurate. In some larger cases, this can have a >99.99% accuracy.

Input and Output

You will take one number in any convienent format.

Examples

Format: Input -> Output

2 -&gt; 1.50
5 -&gt; 2.25
15 -&gt; 4.00
19 -&gt; 4.37               // actually 4.37       + 1/200
27 -&gt; 5.20
39 -&gt; 6.25
47 -&gt; 6.91               // actually 6.91       + 1/300
57 -&gt; 7.57               // actually 7.57       + 1/700
2612 -&gt; 51.10            // actually 51.10      + 2/255
643545345 -&gt; 25368.19    // actually 25,368.19  + 250,000,000/45,113,102,859
35235234236 -&gt; 187710.50 // actually 187,710.50 + 500,000,000/77,374,278,481


Specifications


Your output must be rounded to at least the nearest hundredth (ie. if the answer is 47.2851, you may output 47.29)
Your output does not have to have following zeros and a decimal point if the answer is a whole number (ie. 125.00 can be outputted as 125 and 125.0, too)
You do not have to support any numbers below 1.
You do not have to support non-integer inputs. (ie. 1.52 etc...)


Rules

Standard Loopholes are forbidden.

This is a code-golf, so shortest answer in bytes wins.
Python 2, 47 ... 36 bytes

-3 bytes thanks to @JungHwanMin
-1 byte thanks to @HyperNeutrino
-2 bytes thanks to @JonathanFrech
-3 bytes thanks to @OlivierGrégoire



def f(x):s=int(x**.5);print(x/s+s)/2


Try it online!