4u^2+40u+100=0

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Solution for 4u^2+40u+100=0 equation:



4u^2+40u+100=0
a = 4; b = 40; c = +100;
Δ = b2-4ac
Δ = 402-4·4·100
Δ = 0
Delta is equal to zero, so there is only one solution to the equation
Stosujemy wzór:
$u=\frac{-b}{2a}=\frac{-40}{8}=-5$

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