d(1/6)+(2/3)=(1/4)(d-2)

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Solution for d(1/6)+(2/3)=(1/4)(d-2) equation: 



d(1/6)+(2/3)=(1/4)(d-2)

We move all terms to the left:

d(1/6)+(2/3)-((1/4)(d-2))=0



Domain of the equation: 4)(d-2))!=0

d∈R



We add all the numbers together, and all the variables

d(+1/6)-((+1/4)(d-2))+(+2/3)=0

We multiply parentheses

d^2-((+1/4)(d-2))+(+2/3)=0

We get rid of parentheses

d^2-((+1/4)(d-2))+2/3=0

We multiply parentheses ..

d^2-((+d^2+1/4*-2))+2/3=0

We calculate fractions


d^2+(-((+d^2+1*3)/()+()/()=0



We calculate terms in parentheses: +(-((+d^2+1*3)/()+()/(), so:

-((+d^2+1*3)/()+()/(

We can not solve this equation

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