Determine all the roots of f (t) = 9t^3-18t^2+6t

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