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5x^2+40x+90=10
We move all terms to the left:
5x^2+40x+90-(10)=0
We add all the numbers together, and all the variables
5x^2+40x+80=0
a = 5; b = 40; c = +80;
Δ = b2-4ac
Δ = 402-4·5·80
Δ = 0
Delta is equal to zero, so there is only one solution to the equation
Stosujemy wzór:$x=\frac{-b}{2a}=\frac{-40}{10}=-4$
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