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(3+z)2z=3z(z+2)
We move all terms to the left:
(3+z)2z-(3z(z+2))=0
We add all the numbers together, and all the variables
(z+3)2z-(3z(z+2))=0
We multiply parentheses
2z^2+6z-(3z(z+2))=0
We calculate terms in parentheses: -(3z(z+2)), so:We get rid of parentheses
3z(z+2)
We multiply parentheses
3z^2+6z
Back to the equation:
-(3z^2+6z)
2z^2-3z^2+6z-6z=0
We add all the numbers together, and all the variables
-1z^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:$z=\frac{-b}{2a}=\frac{0}{-2}=0$
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