8x=4x^2-1
solve by completing the square

Get the x terms by themselves on one side and a constant on the other side of the equal sign...

4x^2-8x=1  make the leading coefficient equal to one...

x^2-2x=1/4  now halve the linear coefficient, -2 in this case, square it, and add that value to both sides of the equation...-2/2=-1, -1^2=1 so

x^2-2x+1=1+1/4

x^2-2x+1=5/4  now the left side is a perfect square...

(x-1)^2=5/4  take the square root of both sides...

x-1=±√(5/4)  add 1 to both sides

x=1±√(5/4)