(4/x^2+7x+12)+((3x+18)/x+3)=(1/x^2+7x+12)

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Solution for (4/x^2+7x+12)+((3x+18)/x+3)=(1/x^2+7x+12) equation: 



(4/x^2+7x+12)+((3x+18)/x+3)=(1/x^2+7x+12)

We move all terms to the left:

(4/x^2+7x+12)+((3x+18)/x+3)-((1/x^2+7x+12))=0



Domain of the equation: x^2+7x+12)!=0

x∈R





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

x∈R





Domain of the equation: x^2+7x+12))!=0

x∈R



We get rid of parentheses

4/x^2+7x+((3x+18)/x+3)-((1/x^2+7x+12))+12=0

We calculate fractions

We do not support expression: x^3

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