(z-4)(z-5)=(z+4)(z-3)-8

Solution for (z-4)(z-5)=(z+4)(z-3)-8 equation:

(z-4)(z-5)=(z+4)(z-3)-8

We move all terms to the left:

(z-4)(z-5)-((z+4)(z-3)-8)=0

We multiply parentheses ..

(+z^2-5z-4z+20)-((z+4)(z-3)-8)=0


We calculate terms in parentheses: -((z+4)(z-3)-8), so:

(z+4)(z-3)-8

We multiply parentheses ..

(+z^2-3z+4z-12)-8

We get rid of parentheses

z^2-3z+4z-12-8

We add all the numbers together, and all the variables

z^2+z-20

Back to the equation:

-(z^2+z-20)


We get rid of parentheses

z^2-z^2-5z-4z-z+20+20=0

We add all the numbers together, and all the variables

-10z+40=0

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

-10z=-40

z=-40/-10

z=+4