(y/3)+11=(7/2)-3

Solution for (y/3)+11=(7/2)-3 equation:

(y/3)+11=(7/2)-3

We move all terms to the left:

(y/3)+11-((7/2)-3)=0

We add all the numbers together, and all the variables

(+y/3)+11-((+7/2)-3)=0

We get rid of parentheses

y/3+11-((+7/2)-3)=0

We calculate fractions


2y^2/()+11+()/()=0

We add all the numbers together, and all the variables

2y^2/()+12=0

We multiply all the terms by the denominator


2y^2+12*()=0

We add all the numbers together, and all the variables

2y^2=0

a = 2; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·2·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$y=\frac{-b}{2a}=\frac{0}{4}=0$