(4+2d)/4(d-8)(d-8)=7/16

Solution for (4+2d)/4(d-8)(d-8)=7/16 equation:

(4+2d)/4(d-8)(d-8)=7/16

We move all terms to the left:

(4+2d)/4(d-8)(d-8)-(7/16)=0


Domain of the equation: 4(d-8)(d-8)!=0

d∈R


We add all the numbers together, and all the variables

(2d+4)/4(d-8)(d-8)-(+7/16)=0

We get rid of parentheses

(2d+4)/4(d-8)(d-8)-7/16=0

We multiply parentheses ..

(2d+4)/4(+d^2-8d-8d+64)-7/16=0

We calculate fractions


(32d+64)/(64d^2-1024d+4096)+(-28d(+)/(64d^2-1024d+4096)=0


We calculate terms in parentheses: +(-28d(+)/(64d^2-1024d+4096), so:

-28d(+)/(64d^2-1024d+4096

We add all the numbers together, and all the variables

-28d0/(64d^2-1024d+4096

We multiply all the terms by the denominator


-28d0

We add all the numbers together, and all the variables

-28d

Back to the equation:

+(-28d)


We get rid of parentheses

(32d+64)/(64d^2-1024d+4096)-28d=0

We multiply all the terms by the denominator


(32d+64)-28d*(64d^2-1024d+4096)=0

We multiply parentheses

-1792d^3+28672d^2+(32d+64)-114688d=0

We get rid of parentheses

-1792d^3+28672d^2+32d-114688d+64=0

We do not support edpression: d^3