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=3E^2+6E^2
We move all terms to the left:
-(3E^2+6E^2)=0
We get rid of parentheses
-3E^2-6E^2=0
We add all the numbers together, and all the variables
-9E^2=0
a = -9; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·(-9)·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation
Stosujemy wzór:$E=\frac{-b}{2a}=\frac{0}{-18}=0$
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