Answer
$2012$.
Work Step by Step
Given $y=4.065x^2+370.1x+3450$, for $y=8605$, we have $8605=4.065x^2+370.1x+3450$ or $4.065x^2+370.1x-5155=0$ which can be solved using the quadratic formula (discard the negative answer) to get $x\approx12$ meaning year $2012$.
