Respuesta :

ghanami
hello :
f(t) = 0 :    3t(3t²-6t+2) = 0
(t=0) or (3t²-6t+2 = 0 )
Δ = b²-4ac    .... a=3       b=-6      c=2
Δ = (-6)²-4(3)(2) =36-24 =12 =(2√3)²
x1 =( 6-2√3)/6     ....    x2 =( 6+2√3)/6
the roots are : 0 , ( 6-2√3)/6 , ( 6+2√3)/6

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