1/3y+2.3=3y+2

Solution for 1/3y+2.3=3y+2 equation:

1/3y+2.3=3y+2

We move all terms to the left:

1/3y+2.3-(3y+2)=0


Domain of the equation: 3y!=0

y!=0/3

y!=0

y∈R


We get rid of parentheses

1/3y-3y-2+2.3=0

We multiply all the terms by the denominator


-3y*3y-2*3y+(2.3)*3y+1=0

We multiply parentheses

-3y*3y-2*3y+6.9y+1=0

Wy multiply elements

-9y^2-6y+6.9y+1=0

We add all the numbers together, and all the variables

-9y^2+0.9y+1=0

a = -9; b = 0.9; c = +1;
Δ = b2-4ac
Δ = 0.92-4·(-9)·1
Δ = 36.81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0.9)-\sqrt{36.81}}{2*-9}=\frac{-0.9-\sqrt{36.81}}{-18}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.9)+\sqrt{36.81}}{2*-9}=\frac{-0.9+\sqrt{36.81}}{-18}
$