25t2-20t+4=0

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Solution for 25t2-20t+4=0 equation: 



25t^2-20t+4=0

a = 25; b = -20; c = +4;
Δ = b2-4ac
Δ = -202-4·25·4
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$t=\frac{-b}{2a}=\frac{20}{50}=2/5$

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