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r^2+13+12=
We move all terms to the left:
r^2+13+12-()=0
We add all the numbers together, and all the variables
r^2=0
a = 1; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·1·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation
Stosujemy wzór:$r=\frac{-b}{2a}=\frac{0}{2}=0$
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