1/x+5-3x-7/3x^2+19x+20=1/3x+4

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Solution for 1/x+5-3x-7/3x^2+19x+20=1/3x+4 equation: 



1/x+5-3x-7/3x^2+19x+20=1/3x+4

We move all terms to the left:

1/x+5-3x-7/3x^2+19x+20-(1/3x+4)=0



Domain of the equation: x!=0

x∈R





Domain of the equation: 3x^2!=0

x^2!=0/3

x^2!=√0

x!=0

x∈R





Domain of the equation: 3x+4)!=0

x∈R



We add all the numbers together, and all the variables

16x+1/x-7/3x^2-(1/3x+4)+25=0

We get rid of parentheses

16x+1/x-7/3x^2-1/3x-4+25=0

We calculate fractions

We do not support expression: x^4

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