y^2=4-x^2
y=(4-x^2)^0.5 i suppose that it is equal to you second equation also.
you must consider the right hand side of you equation in brakets when taking roots/squares. hope that helps
(you must take the posotive and negative of (4-x^2)^0.5)
Hi RB89,
If you're wanting to solve for y, you will need to take the square root of both sides of the equation.
And that's about it.
Now, if you had something like: , you could get to because the right side of the equation was a perfect square. That wasn't the case in your example.