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t^2-3/5t-1.5/5=0
Domain of the equation: 5t!=0We calculate fractions
t!=0/5
t!=0
t∈R
t^2=0
a = 1; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·1·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation
Stosujemy wzór:$t=\frac{-b}{2a}=\frac{0}{2}=0$
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