Ask Me Help Desk - Solve using 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.