20x+100+x^2=0

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Solution for 20x+100+x^2=0 equation: 



20x+100+x^2=0

a = 1; b = 20; c = +100;
Δ = b2-4ac
Δ = 202-4·1·100
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$x=\frac{-b}{2a}=\frac{-20}{2}=-10$

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