(7/6)d+(4/3)=(-1/3)

Solution for (7/6)d+(4/3)=(-1/3) equation:

(7/6)d+(4/3)=(-1/3)

We move all terms to the left:

(7/6)d+(4/3)-((-1/3))=0


Domain of the equation: 6)d!=0

d!=0/1

d!=0

d∈R


We add all the numbers together, and all the variables

(+7/6)d+(+4/3)-((-1/3))=0

We multiply parentheses

7d^2+(+4/3)-((-1/3))=0

We get rid of parentheses

7d^2+4/3-((-1/3))=0

We calculate fractions


7d^2=0

a = 7; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·7·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$d=\frac{-b}{2a}=\frac{0}{14}=0$