10x^2+20x=-10

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Solution for 10x^2+20x=-10 equation: 



10x^2+20x=-10

We move all terms to the left:

10x^2+20x-(-10)=0

We add all the numbers together, and all the variables

10x^2+20x+10=0

a = 10; b = 20; c = +10;
Δ = b2-4ac
Δ = 202-4·10·10
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$x=\frac{-b}{2a}=\frac{-20}{20}=-1$

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