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

Solution for (8-z)(3z-5)=0 equation:

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

We add all the numbers together, and all the variables

(-1z+8)(3z-5)=0

We multiply parentheses ..

(-3z^2+5z+24z-40)=0

We get rid of parentheses

-3z^2+5z+24z-40=0

We add all the numbers together, and all the variables

-3z^2+29z-40=0

a = -3; b = 29; c = -40;
Δ = b2-4ac
Δ = 292-4·(-3)·(-40)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-19}{2*-3}=\frac{-48}{-6}
=+8
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+19}{2*-3}=\frac{-10}{-6}
=1+2/3
$