(1/3)y+2.3=3y-3

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Solution for (1/3)y+2.3=3y-3 equation: 



(1/3)y+2.3=3y-3

We move all terms to the left:

(1/3)y+2.3-(3y-3)=0



Domain of the equation: 3)y!=0

y!=0/1

y!=0

y∈R



We add all the numbers together, and all the variables

(+1/3)y-(3y-3)+2.3=0

We multiply parentheses

y^2-(3y-3)+2.3=0

We get rid of parentheses

y^2-3y+3+2.3=0

We add all the numbers together, and all the variables

y^2-3y+5.3=0

a = 1; b = -3; c = +5.3;
Δ = b2-4ac
Δ = -32-4·1·5.3
Δ = -12.2
Delta is less than zero, so there is no solution for the equation

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