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16t^2-8t+1=0
a = 16; b = -8; c = +1;
Δ = b2-4ac
Δ = -82-4·16·1
Δ = 0
Delta is equal to zero, so there is only one solution to the equation
Stosujemy wzór:$t=\frac{-b}{2a}=\frac{8}{32}=1/4$
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