y+y+20+1/3y=132/3

Solution for y+y+20+1/3y=132/3 equation:

y+y+20+1/3y=132/3

We move all terms to the left:

y+y+20+1/3y-(132/3)=0


Domain of the equation: 3y!=0

y!=0/3

y!=0

y∈R


We add all the numbers together, and all the variables

y+y+1/3y+20-44=0

We add all the numbers together, and all the variables

2y+1/3y-24=0

We multiply all the terms by the denominator


2y*3y-24*3y+1=0

Wy multiply elements

6y^2-72y+1=0

a = 6; b = -72; c = +1;
Δ = b2-4ac
Δ = -722-4·6·1
Δ = 5160
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{5160}=\sqrt{4*1290}=\sqrt{4}*\sqrt{1290}=2\sqrt{1290}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-72)-2\sqrt{1290}}{2*6}=\frac{72-2\sqrt{1290}}{12}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-72)+2\sqrt{1290}}{2*6}=\frac{72+2\sqrt{1290}}{12}
$