28=-16t^2+40t+3

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Solution for 28=-16t^2+40t+3 equation: 



28=-16t^2+40t+3

We move all terms to the left:

28-(-16t^2+40t+3)=0

We get rid of parentheses

16t^2-40t-3+28=0

We add all the numbers together, and all the variables

16t^2-40t+25=0

a = 16; b = -40; c = +25;
Δ = b2-4ac
Δ = -402-4·16·25
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$t=\frac{-b}{2a}=\frac{40}{32}=1+1/4$

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