3(t+2)+2(t-3)=4(t+2)+5(t+4)

Solution for 3(t+2)+2(t-3)=4(t+2)+5(t+4) equation:

3(t+2)+2(t-3)=4(t+2)+5(t+4)

We move all terms to the left:

3(t+2)+2(t-3)-(4(t+2)+5(t+4))=0

We multiply parentheses

3t+2t-(4(t+2)+5(t+4))+6-6=0


We calculate terms in parentheses: -(4(t+2)+5(t+4)), so:

4(t+2)+5(t+4)

We multiply parentheses

4t+5t+8+20

We add all the numbers together, and all the variables

9t+28

Back to the equation:

-(9t+28)


We add all the numbers together, and all the variables

5t-(9t+28)=0

We get rid of parentheses

5t-9t-28=0

We add all the numbers together, and all the variables

-4t-28=0

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

-4t=28

t=28/-4

t=-7