-50t^2+100t-50=0

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Solution for -50t^2+100t-50=0 equation: 



-50t^2+100t-50=0

a = -50; b = 100; c = -50;
Δ = b2-4ac
Δ = 1002-4·(-50)·(-50)
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$t=\frac{-b}{2a}=\frac{-100}{-100}=1$

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