(-2/3)z-2=10

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Solution for (-2/3)z-2=10 equation: 



(-2/3)z-2=10

We move all terms to the left:

(-2/3)z-2-(10)=0



Domain of the equation: 3)z!=0

z!=0/1

z!=0

z∈R



We add all the numbers together, and all the variables

(-2/3)z-12=0

We multiply parentheses

-2z^2-12=0

a = -2; b = 0; c = -12;
Δ = b2-4ac
Δ = 02-4·(-2)·(-12)
Δ = -96
Delta is less than zero, so there is no solution for the equation

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