13(z+4)-2(z-3)=6(z-2)+4(z-2)

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Solution for 13(z+4)-2(z-3)=6(z-2)+4(z-2) equation: 



13(z+4)-2(z-3)=6(z-2)+4(z-2)

We move all terms to the left:

13(z+4)-2(z-3)-(6(z-2)+4(z-2))=0

We multiply parentheses

13z-2z-(6(z-2)+4(z-2))+52+6=0



We calculate terms in parentheses: -(6(z-2)+4(z-2)), so:

6(z-2)+4(z-2)

We multiply parentheses

6z+4z-12-8

We add all the numbers together, and all the variables

10z-20

Back to the equation:

-(10z-20)



We add all the numbers together, and all the variables

11z-(10z-20)+58=0

We get rid of parentheses

11z-10z+20+58=0

We add all the numbers together, and all the variables

z+78=0

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

z=-78

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