log(2)α=t
t^2+2t-3=0; D=4+12=16
t1=(-2+4)/2=1; log(2)α=1; α1=2^1=2
t2=(-2-4)/2=-3; log(2)α=-3; α2=2^(-3)=1/8
Vпо.теч=9/(3/4)=12 км/ч.