(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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