49f^2+140f+100=0

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Solution for 49f^2+140f+100=0 equation: 



49f^2+140f+100=0

a = 49; b = 140; c = +100;
Δ = b2-4ac
Δ = 1402-4·49·100
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$f=\frac{-b}{2a}=\frac{-140}{98}=-1+3/7$

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