-5t^2+20t-20=0

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Solution for -5t^2+20t-20=0 equation: 



-5t^2+20t-20=0

a = -5; b = 20; c = -20;
Δ = b2-4ac
Δ = 202-4·(-5)·(-20)
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$t=\frac{-b}{2a}=\frac{-20}{-10}=+2$

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