20=-16t^2+32t+4

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Solution for 20=-16t^2+32t+4 equation: 



20=-16t^2+32t+4

We move all terms to the left:

20-(-16t^2+32t+4)=0

We get rid of parentheses

16t^2-32t-4+20=0

We add all the numbers together, and all the variables

16t^2-32t+16=0

a = 16; b = -32; c = +16;
Δ = b2-4ac
Δ = -322-4·16·16
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$t=\frac{-b}{2a}=\frac{32}{32}=1$

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