2(t+2)-5=5(t-4)+13

Solution for 2(t+2)-5=5(t-4)+13 equation:

2(t+2)-5=5(t-4)+13

We move all terms to the left:

2(t+2)-5-(5(t-4)+13)=0

We multiply parentheses

2t-(5(t-4)+13)+4-5=0


We calculate terms in parentheses: -(5(t-4)+13), so:

5(t-4)+13

We multiply parentheses

5t-20+13

We add all the numbers together, and all the variables

5t-7

Back to the equation:

-(5t-7)


We add all the numbers together, and all the variables

2t-(5t-7)-1=0

We get rid of parentheses

2t-5t+7-1=0

We add all the numbers together, and all the variables

-3t+6=0

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

-3t=-6

t=-6/-3

t=+2