(3/8)f+(1/2)=6((1/16)f-3)

Solution for (3/8)f+(1/2)=6((1/16)f-3) equation:

(3/8)f+(1/2)=6((1/16)f-3)

We move all terms to the left:

(3/8)f+(1/2)-(6((1/16)f-3))=0


Domain of the equation: 8)f!=0

f!=0/1

f!=0

f∈R



Domain of the equation: 16)f-3))!=0

f!=0/1

f!=0

f∈R


We add all the numbers together, and all the variables

(+3/8)f-(6((+1/16)f-3))+(+1/2)=0

We multiply parentheses

3f^2-(6((+1/16)f-3))+(+1/2)=0

We get rid of parentheses

3f^2-(6((+1/16)f-3))+1/2=0

We calculate fractions


3f^2+()/16f-3))*2)+16f/16f-3))*2)=0

We calculate fractions


3f^2+(()*16f)/256f^2+(-3))*2)+16f*16f)/256f^2=0

We multiply all the terms by the denominator


3f^2*256f^2+(()*16f)+(-3))*2)+16f*16f)=0


We calculate terms in parentheses: +(()*16f), so:

()*16f


Wy multiply elements

768f^4+(()*16f)+(-3))*2)+16f*16f)=0

We get rid of parentheses

768f^4+(()*16f)-3))*2)+16f*16f=0


We calculate terms in parentheses: +(()*16f), so:

()*16f


We do not support efpression: f^4