(2/3)(9c-3)=22

Solution for (2/3)(9c-3)=22 equation:

(2/3)(9c-3)=22

We move all terms to the left:

(2/3)(9c-3)-(22)=0


Domain of the equation: 3)(9c-3)!=0

c∈R


We add all the numbers together, and all the variables

(+2/3)(9c-3)-22=0

We multiply parentheses ..

(+18c^2+2/3*-3)-22=0

We multiply all the terms by the denominator


(+18c^2+2-22*3*-3)=0

We get rid of parentheses

18c^2+2-3-22*3*=0

We add all the numbers together, and all the variables

18c^2=0

a = 18; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·18·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$c=\frac{-b}{2a}=\frac{0}{36}=0$