Hi,

You need to rearrange this equation so that $x$ is alone on the left side and the right side has no $x.$ I would first multiply both sides by 2 to eliminate the fraction. Thus, multiplying both sides by 2 the equation becomes

\[2y = (x-1)^{2}.\]

If you now take the square of both sides you get

\[\sqrt{2y} = \sqrt {\left( x-1 \right)^2} = \pm (x-1).\]

Hence I get two equations

\[\sqrt{2y} = x-1 \mbox{ and } \sqrt{2y} = -(x-1) = 1-x.\]

Hence either

\[x = \sqrt{2y} + 1\]

or

\[x = 1 - \sqrt{2y}.\]

Penny