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k^2+6k+6k+36=0
We add all the numbers together, and all the variables
k^2+12k+36=0
a = 1; b = 12; c = +36;
Δ = b2-4ac
Δ = 122-4·1·36
Δ = 0
Delta is equal to zero, so there is only one solution to the equation
Stosujemy wzór:$k=\frac{-b}{2a}=\frac{-12}{2}=-6$
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