Need detailed steps and solutions using the indian method.

Thanks

x^2-2x-3=0

First, move the constant term to the right side of the equation:

x^2-2x=3

Now multiply everything by four times the original coefficient of the x^2 term (i.e. 4 x 1 = 4):

4x^2-8x=12

Next, add the square of the original coefficient of x to both sides (i.e. [-2]^2=4):

4x^2-8x+4=16

This makes the left side of the equation a perfect square. So now you can take the square root of both sides:

\sqrt{4x^2-8x+4}=\sqrt{16}

\sqrt{(2x-2)^2}=\sqrt{16}

2x-2=\pm 4

x=-1 or x=3