(2/3)+(y/4)=(28/6)

Solution for (2/3)+(y/4)=(28/6) equation:

(2/3)+(y/4)=(28/6)

We move all terms to the left:

(2/3)+(y/4)-((28/6))=0

We add all the numbers together, and all the variables

(+y/4)+(+2/3)-((+28/6))=0

We get rid of parentheses

y/4+2/3-((+28/6))=0

We calculate fractions


108y^2/()+()/()+()/()=0

We add all the numbers together, and all the variables

108y^2/()+2=0

We multiply all the terms by the denominator


108y^2+2*()=0

We add all the numbers together, and all the variables

108y^2=0

a = 108; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·108·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$y=\frac{-b}{2a}=\frac{0}{216}=0$