(i) Given that \(y=2x^2-4x-7\), write \(y\) in the form \(a(x-b)^2+c\), where \(a\), \(b\) and \(c\) are constants.

(ii) Hence write down the minimum value of \(y\) and the value of \(x\) at which it occurs.

(iii) Using your answer to part (i), solve

\(2p-4\sqrt p-7=0,\)

giving your answer correct to 2 decimal places.