The number, \(B\), of a certain type of bacteria at time \(t\) days is described by \(B=200e^{2t}+800e^{-2t}\).

(i) Find \(B\) when \(t=0\).

(ii) At the instant when \(\frac{dB}{dt}=1200\), show that \(e^{4t}-3e^{2t}-4=0\).

(iii) Using the substitution \(u=e^{2t}\), or otherwise, solve \(e^{4t}-3e^{2t}-4=0\).