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20t-5t^2=20
We move all terms to the left:
20t-5t^2-(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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