8(t+2)-3*(t-4)=6(t-7)+8

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Solution for 8(t+2)-3*(t-4)=6(t-7)+8 equation: 



8(t+2)-3(t-4)=6(t-7)+8

We move all terms to the left:

8(t+2)-3(t-4)-(6(t-7)+8)=0

We multiply parentheses

8t-3t-(6(t-7)+8)+16+12=0



We calculate terms in parentheses: -(6(t-7)+8), so:

6(t-7)+8

We multiply parentheses

6t-42+8

We add all the numbers together, and all the variables

6t-34

Back to the equation:

-(6t-34)



We add all the numbers together, and all the variables

5t-(6t-34)+28=0

We get rid of parentheses

5t-6t+34+28=0

We add all the numbers together, and all the variables

-1t+62=0

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

-t=-62

t=-62/-1

t=+62

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