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)

Isolate the 1: 4x^2-8x=1

Since "a" is a perfect square use the formula b^2/4a: (-8) ^2/4 (4) = 4

Add 4 into both sides of the equation:

4x^2-8x+4=1+4

Ur new equation should be:

4x^2 - 8x + 4 = 5

Factor: (2x-2) ^2=5

Solve for x by getting the square root of 5 to get rid of the square on the (2x-2) ^2 side and you would get:

2x-2 = + - 2.2

solve for x: x = 2.1 x = - 0.1