(2/3)(3t-6)+1=5

Solution for (2/3)(3t-6)+1=5 equation:

(2/3)(3t-6)+1=5

We move all terms to the left:

(2/3)(3t-6)+1-(5)=0


Domain of the equation: 3)(3t-6)!=0

t∈R


We add all the numbers together, and all the variables

(+2/3)(3t-6)+1-5=0

We add all the numbers together, and all the variables

(+2/3)(3t-6)-4=0

We multiply parentheses ..

(+6t^2+2/3*-6)-4=0

We multiply all the terms by the denominator


(+6t^2+2-4*3*-6)=0

We get rid of parentheses

6t^2+2-6-4*3*=0

We add all the numbers together, and all the variables

6t^2=0

a = 6; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·6·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$t=\frac{-b}{2a}=\frac{0}{12}=0$