20=-5t^2+20t

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



20=-5t^2+20t

We move all terms to the left:

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

We get rid of parentheses

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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