(8+c)(c-3)-(6+c)(c-4)=0

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Solution for (8+c)(c-3)-(6+c)(c-4)=0 equation:



(8+c)(c-3)-(6+c)(c-4)=0
We add all the numbers together, and all the variables
(c+8)(c-3)-(c+6)(c-4)=0
We multiply parentheses ..
(+c^2-3c+8c-24)-(c+6)(c-4)=0
We get rid of parentheses
c^2-3c+8c-(c+6)(c-4)-24=0
We multiply parentheses ..
c^2-(+c^2-4c+6c-24)-3c+8c-24=0
We add all the numbers together, and all the variables
c^2-(+c^2-4c+6c-24)+5c-24=0
We get rid of parentheses
c^2-c^2+4c-6c+5c+24-24=0
We add all the numbers together, and all the variables
3c=0
c=0/3
c=0

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