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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-3z^2+9z+5z-15)=0

We get rid of parentheses

-3z^2+9z+5z-15=0

We add all the numbers together, and all the variables

-3z^2+14z-15=0

a = -3; b = 14; c = -15;
Δ = b2-4ac
Δ = 142-4·(-3)·(-15)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-4}{2*-3}=\frac{-18}{-6}
=+3
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+4}{2*-3}=\frac{-10}{-6}
=1+2/3
$