3/5(15d+20)-(1/6)(18d-12)=38

Solution for 3/5(15d+20)-(1/6)(18d-12)=38 equation:

3/5(15d+20)-(1/6)(18d-12)=38

We move all terms to the left:

3/5(15d+20)-(1/6)(18d-12)-(38)=0


Domain of the equation: 5(15d+20)!=0

d∈R



Domain of the equation: 6)(18d-12)!=0

d∈R


We add all the numbers together, and all the variables

3/5(15d+20)-(+1/6)(18d-12)-38=0

We multiply parentheses ..

-(+18d^2+1/6*-12)+3/5(15d+20)-38=0

We calculate fractions


(-(18d^2+5d1)/15d-12)*5()+()/15d-38-12)*5()=0


We calculate terms in parentheses: +(-(18d^2+5d1)/15d-12)*5(), so:

-(18d^2+5d1)/15d-12)*5(

We add all the numbers together, and all the variables

-(18d^2+5d1)/15d

We multiply all the terms by the denominator


-(18d^2+5d1)

We get rid of parentheses

-18d^2-5d1

We add all the numbers together, and all the variables

-18d^2-5d

Back to the equation:

+(-18d^2-5d)


We add all the numbers together, and all the variables

(-18d^2-5d)+()/15d=0

We get rid of parentheses

-18d^2-5d+()/15d=0

We multiply all the terms by the denominator


-18d^2*15d-5d*15d+()=0

We add all the numbers together, and all the variables

-18d^2*15d-5d*15d=0

Wy multiply elements

-270d^3-75d^2=0

We do not support edpression: d^3