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(2+3)(D2+2D+1)=0
We add all the numbers together, and all the variables
5(+D^2+2D+1)=0
We multiply parentheses
5D^2+10D+5=0
a = 5; b = 10; c = +5;
Δ = b2-4ac
Δ = 102-4·5·5
Δ = 0
Delta is equal to zero, so there is only one solution to the equation
Stosujemy wzór:$D=\frac{-b}{2a}=\frac{-10}{10}=-1$
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