5(2d-5)-4=4(d+3)+13

Solution for 5(2d-5)-4=4(d+3)+13 equation:

5(2d-5)-4=4(d+3)+13

We move all terms to the left:

5(2d-5)-4-(4(d+3)+13)=0

We multiply parentheses

10d-(4(d+3)+13)-25-4=0


We calculate terms in parentheses: -(4(d+3)+13), so:

4(d+3)+13

We multiply parentheses

4d+12+13

We add all the numbers together, and all the variables

4d+25

Back to the equation:

-(4d+25)


We add all the numbers together, and all the variables

10d-(4d+25)-29=0

We get rid of parentheses

10d-4d-25-29=0

We add all the numbers together, and all the variables

6d-54=0

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

6d=54

d=54/6

d=9