2t^2+32t+128=0

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Solution for 2t^2+32t+128=0 equation: 



2t^2+32t+128=0

a = 2; b = 32; c = +128;
Δ = b2-4ac
Δ = 322-4·2·128
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$t=\frac{-b}{2a}=\frac{-32}{4}=-8$

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