((w^2-10)/(w+2))+w-4=w-3

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Solution for ((w^2-10)/(w+2))+w-4=w-3 equation: 



((w^2-10)/(w+2))+w-4=w-3

We move all terms to the left:

((w^2-10)/(w+2))+w-4-(w-3)=0



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

w∈R



We add all the numbers together, and all the variables

w+((w^2-10)/(w+2))-(w-3)-4=0

We get rid of parentheses

w+((w^2-10)/(w+2))-w+3-4=0

We multiply all the terms by the denominator


w*(w+2))+((w^2-10)-w*(w+2))+3*(w+2))-4*(w+2))=0

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