(k-7)(k-4)-6=(k+2)(k-8)

Solution for (k-7)(k-4)-6=(k+2)(k-8) equation:

(k-7)(k-4)-6=(k+2)(k-8)

We move all terms to the left:

(k-7)(k-4)-6-((k+2)(k-8))=0

We multiply parentheses ..

(+k^2-4k-7k+28)-((k+2)(k-8))-6=0


We calculate terms in parentheses: -((k+2)(k-8)), so:

(k+2)(k-8)

We multiply parentheses ..

(+k^2-8k+2k-16)

We get rid of parentheses

k^2-8k+2k-16

We add all the numbers together, and all the variables

k^2-6k-16

Back to the equation:

-(k^2-6k-16)


We get rid of parentheses

k^2-k^2-4k-7k+6k+28+16-6=0

We add all the numbers together, and all the variables

-5k+38=0

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

-5k=-38

k=-38/-5

k=7+3/5