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125=-5t^2+50t
We move all terms to the left:
125-(-5t^2+50t)=0
We get rid of parentheses
5t^2-50t+125=0
a = 5; b = -50; c = +125;
Δ = b2-4ac
Δ = -502-4·5·125
Δ = 0
Delta is equal to zero, so there is only one solution to the equation
Stosujemy wzór:$t=\frac{-b}{2a}=\frac{50}{10}=5$
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