t^2+28t+196=0

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Solution for t^2+28t+196=0 equation: 



t^2+28t+196=0

a = 1; b = 28; c = +196;
Δ = b2-4ac
Δ = 282-4·1·196
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$t=\frac{-b}{2a}=\frac{-28}{2}=-14$

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