Calculating the peak output voltage from RMS power of an audio amplifier
With some assumptions (pure sine wave, resistive load, etc.) you can simply apply your formula. 

$$ 50 = \frac{V^2}{8} $$

leads to

$$ V = \sqrt{400} $$

hence \$V = 20\$. This is the DC voltage that puts 50&nbsp;W in an 8&nbsp;&Omega; load.

For an AC sine wave the peak (with respect to the average) is \$\sqrt{2}\$ times this value, or approximately 28&nbsp;V. These peaks appear both in positive and negative direction, so the peak-to-peak voltage is twice this value, approximately 56&nbsp;V. 

This is the minimal DC supply a capacitor-coupled push-pull amplifier stage would need. A direct-coupled push-pull amplifier would need one positive and one negative 28&nbsp;V supplies, a bridge amplifier would need one 28&nbsp;V supply. (In each case, more is needed due to losses, margins, etc.)