1x^2+-20x=-100

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Solution for 1x^2+-20x=-100 equation: 



1x^2+-20x=-100

We move all terms to the left:

1x^2+-20x-(-100)=0

We add all the numbers together, and all the variables

x^2-20x+100=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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