Exponents across = signs Solve for $V$:
$$ R=V^2/(2g) $$ 
$$ R\cdot (2g) = V^2 $$
$$ \sqrt{R\cdot(2g)} = V $$
$$ V = \sqrt{R\cdot(2g)} $$
Is this correct? Or should I move the exponent in the first step to get:
$$ V = \sqrt{R}\cdot (2g) $$
Thanks
 A: Both approaches are more or less correct, but your second approach is wrongly executed. There is an issue with your execution of your first approach.
First an minor issue is that you should know that $2g\ne 0$, but since you divide by that in the equation one can assume that's true.
Second is the step when you go from $R(2g) = V^2$ to $\sqrt{R(2g)} =V$ you're assuming that $V$ should be non-negative. The domain problem may require this, but in general one would solve for both solutions $\pm\sqrt{R(2g)} = V$.
Third you have to know that $R(2g)\ge 0$, or you would not have any real solutions.
Your second approach is also correct, but incorrectly executed. The correct execution is:
$$R = V^2/(2g)$$
$$\pm\sqrt{R} = V/\sqrt{2g}$$
$$\pm\sqrt{R}\sqrt{2g}= V$$
$$\pm\sqrt{R(2g)} = V$$
$$V = \pm\sqrt{R(2g)}$$
The first step requires a bit of explaination, moving the exponent means to take square root of the both sides and as mentioned you have to get a $\pm$ to get both solutions.
The second approach has the problem that the third issue becomes a little more complicated since you rely on both $R$ and $2g$ to be non-negative.
