Official solution
Given differential equation is
\[{{dp\left( t \right)} \over {dt}} = 0.5p\left( t \right) - 450\]
\[ \Rightarrow {{dp\left( t \right)} \over {dt}} = {1 \over 2}p\left( t \right) - 450\]
\[ \Rightarrow {{dp\left( t \right)} \over {dt}} = {{p\left( t \right) - 900} \over 2}\]
\[ \Rightarrow 2{{dp\left( t \right)} \over {dt}} = \left[ {900 - p\left( t \right)} \right]\]
\[ \Rightarrow 2{{dp\left( t \right)} \over {900 - p\left( t \right)}} = - dt\]
Integrate both sides, we get
\[\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 2\int {{{dp\left( t \right)} \over {900 - p\left( t \right)}}} = \int {dt} \]
\[\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Let\,\,900 - p\left( t \right) = u\]
\[ \Rightarrow - dp\left( t \right) = du\]
\[\therefore\] \[\,\,\,\,\,\,\] We have,
\[2\int {{{du} \over u}} = \int {dt \Rightarrow 2\,\ln \,u = t + c} \]
\[ \Rightarrow 2\ln \left[ {900 - p\left( t \right)} \right] = t + c\]
\[\,\,\,\,\,\,\,\,\,\,\,\,\,\,\] when \[t = 0,p\left( 0 \right) = 850\]
\[\,\,\,\,\,\,\,\,\,\,\,\,\,\,\] \[2\ln \left( {50} \right) = c\]
\[ \Rightarrow 2\left[ {\ln \left( {{{900 - p\left( t \right)} \over {50}}} \right)} \right] = t\]
\[ \Rightarrow 900 - p\left( t \right) = 50{e^{{t \over 2}}}\]
\[ \Rightarrow p\left( t \right) = 900 - 50{e^{{t \over 2}}}\]
Let \[p\left( {{t_1}} \right) = 0\]
\[0 = 900 - 50{e^{{{{t_1}} \over 2}}}\]
\[\therefore\] \[\,\,\,\,\,\,{t_1} = 2\ln 18\]