8(t+2)-3(t-4)=6(2t-12)-4t

Solution for 8(t+2)-3(t-4)=6(2t-12)-4t equation:

8(t+2)-3(t-4)=6(2t-12)-4t

We move all terms to the left:

8(t+2)-3(t-4)-(6(2t-12)-4t)=0

We multiply parentheses

8t-3t-(6(2t-12)-4t)+16+12=0


We calculate terms in parentheses: -(6(2t-12)-4t), so:

6(2t-12)-4t

We add all the numbers together, and all the variables

-4t+6(2t-12)

We multiply parentheses

-4t+12t-72

We add all the numbers together, and all the variables

8t-72

Back to the equation:

-(8t-72)


We add all the numbers together, and all the variables

5t-(8t-72)+28=0

We get rid of parentheses

5t-8t+72+28=0

We add all the numbers together, and all the variables

-3t+100=0

We move all terms containing t to the left, all other terms to the right

-3t=-100

t=-100/-3

t=33+1/3