(y/6)+5=(1/3)-y

Solution for (y/6)+5=(1/3)-y equation:

(y/6)+5=(1/3)-y

We move all terms to the left:

(y/6)+5-((1/3)-y)=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

(+y/6)-((+1/3)-y)+5=0

We get rid of parentheses

y/6-((+1/3)-y)+5=0

We calculate fractions


3y^2/18y+()/18y+5=0

We multiply all the terms by the denominator


3y^2+5*18y+()=0

We add all the numbers together, and all the variables

3y^2+5*18y=0

Wy multiply elements

3y^2+90y=0

a = 3; b = 90; c = 0;
Δ = b2-4ac
Δ = 902-4·3·0
Δ = 8100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{8100}=90$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(90)-90}{2*3}=\frac{-180}{6}
=-30
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(90)+90}{2*3}=\frac{0}{6}
=0
$