



 
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 \[2y = (x1)^{2}.\] If you now take the square of both sides you get \[\sqrt{2y} = \sqrt {\left( x1 \right)^2} = \pm (x1).\] Hence I get two equations \[\sqrt{2y} = x1 \mbox{ and } \sqrt{2y} = (x1) = 1x.\] Hence either \[x = \sqrt{2y} + 1\] or \[x = 1  \sqrt{2y}.\] Penny 



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