0.1x^2+4x+40=0

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Solution for 0.1x^2+4x+40=0 equation: 



0.1x^2+4x+40=0

a = 0.1; b = 4; c = +40;
Δ = b2-4ac
Δ = 42-4·0.1·40
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$x=\frac{-b}{2a}=\frac{-4}{0.2}=-20$

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