(2t)+8+((1/2)t)-(1t)=3

Solution for (2t)+8+((1/2)t)-(1t)=3 equation:

(2t)+8+((1/2)t)-(1t)=3

We move all terms to the left:

(2t)+8+((1/2)t)-(1t)-(3)=0


Domain of the equation: 2)t)!=0

t!=0/1

t!=0

t∈R


We add all the numbers together, and all the variables

2t+((+1/2)t)-1t+8-3=0

We add all the numbers together, and all the variables

t+((+1/2)t)+5=0

We multiply all the terms by the denominator


t*2)t)+((+5*2)t)+1=0

We add all the numbers together, and all the variables

t*2)t)+(10t)+1=0

We add all the numbers together, and all the variables

t*2)t)+10t+1=0

Wy multiply elements

2t^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:
$t=\frac{-b}{2a}=\frac{0}{4}=0$