(5/w^2+2w-8)+(1/w^2+w-6)=4/(w^2+7w+12)

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Solution for (5/w^2+2w-8)+(1/w^2+w-6)=4/(w^2+7w+12) equation: 



(5/w^2+2w-8)+(1/w^2+w-6)=4/(w^2+7w+12)

We move all terms to the left:

(5/w^2+2w-8)+(1/w^2+w-6)-(4/(w^2+7w+12))=0



Domain of the equation: w^2+2w-8)!=0

w∈R





Domain of the equation: w^2+w-6)!=0

w∈R





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

w∈R



We get rid of parentheses

5/w^2+2w+1/w^2+w-(4/(w^2+7w+12))-8-6=0

We calculate fractions


2w+w+(1*(w^2+7w+12))+5)/(w^2*(w^2+7w+12)))+(-(4*w^2)/(w^2*(w^2+7w+12)))-8-6=0



We calculate terms in parentheses: +(1*(w^2+7w+12)), so:

1*(w^2+7w+12)



We can not solve this equation

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