(d+5)=3+(2d+3)

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

(d+5)=3+(2d+3)

We move all terms to the left:

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

We get rid of parentheses

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


We calculate terms in parentheses: -(3+(2d+3)), so:

3+(2d+3)

determiningTheFunctionDomain
(2d+3)+3

We get rid of parentheses

2d+3+3

We add all the numbers together, and all the variables

2d+6

Back to the equation:

-(2d+6)


We get rid of parentheses

d-2d-6+5=0

We add all the numbers together, and all the variables

-1d-1=0

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

-d=1

d=1/-1

d=-1