x^2-2x-13=0 I need help solving this problem using the Indian Method

First move the -13 to the other side and get x^2-2x=13 .
Since the coefficient of x^2 is one, multiply both sides by 4 and get:
4x^2 - 8x = 52 .
Square the coefficient of the original x term, which was -2 , to get 4 , then add it to both sides and get:
4x^2 -8x +4 = 56 .
Factor 4x^2 -8x +4 and we get:
\Rightarrow (2x - 2)^2 =56 .
Take the square root of both sides and solve for x.
2x - 2 = \pm sqrt{56} .
\Rightarrow 2x = 2 \pm sqrt{56} .

\Rightarrow x = \frac{2 \pm sqrt{56}}{2} .

\Rightarrow x = \frac{2 \pm 2sqrt{14}}{2} .

\Rightarrow x = 1 \pm sqrt{14} .


I had never heard of this "indian method" before, I had to look it up. How this is easier than completing the square I don't know. I think it's for people who see fractions and freak out. Now I feel dirty.