Approximate My Squares

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

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R, 43 bytes 29 bytes

x=scan()
(x/(s=x^.5%/%1)+s)/2

Thanks to @Giuseppe for the new equation and help in golfing of 12 bytes with the integer division solution. By swapping out the function call for scan, I golfed another couple bytes.

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Java (OpenJDK 8), 32 bytes

n->(n/(n=(int)Math.sqrt(n))+n)/2

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Explanations

The code is equivalent to this:

double approx_sqrt(double x) {
  double s = (int)Math.sqrt(x);  // assign the root integer to s
  return (x / s + s) / 2
}

The maths behind:

s + (x - s²) / (2 * s)  =  (2 * s² + x - s²) / (2 * s)
                        =  (x + s²) / (2 * s)
                        =  (x + s²) / s / 2
                        =  ((x + s²) / s) / 2
                        =  (x / s + s² / s) / 2
                        =  (x / s + s) / 2