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12=-3b^2-12b
We move all terms to the left:
12-(-3b^2-12b)=0
We get rid of parentheses
3b^2+12b+12=0
a = 3; b = 12; c = +12;
Δ = b2-4ac
Δ = 122-4·3·12
Δ = 0
Delta is equal to zero, so there is only one solution to the equation
Stosujemy wzór:$b=\frac{-b}{2a}=\frac{-12}{6}=-2$
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