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

Solution for (6-z)(3z-5)=0 equation:

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

We add all the numbers together, and all the variables

(-1z+6)(3z-5)=0

We multiply parentheses ..

(-3z^2+5z+18z-30)=0

We get rid of parentheses

-3z^2+5z+18z-30=0

We add all the numbers together, and all the variables

-3z^2+23z-30=0

a = -3; b = 23; c = -30;
Δ = b2-4ac
Δ = 232-4·(-3)·(-30)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-13}{2*-3}=\frac{-36}{-6}
=+6
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+13}{2*-3}=\frac{-10}{-6}
=1+2/3
$