(.400+2x)^2/(0.004-x)=80

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Solution for (.400+2x)^2/(0.004-x)=80 equation: 



(.400+2x)^2/(0.004-x)=80

We move all terms to the left:

(.400+2x)^2/(0.004-x)-(80)=0



Domain of the equation: (0.004-x)!=0

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

-x!=-0.004

x!=-0.004/-1

x!=0.004/1

x∈R



We add all the numbers together, and all the variables

(2x+0.4)^2/(-1x+0.004)-80=0

We multiply all the terms by the denominator


(2x+0.4)^2-80*(-1x+0.004)=0

We multiply parentheses

(2x+0.4)^2+80x-0.32=0

We add all the numbers together, and all the variables

80x+(2x+0.4)^2-0.32=0

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

80x+(2x+0.4)^2=0.32

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