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

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

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

We multiply parentheses ..

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

We get rid of parentheses

3z^2-18z-5z+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{10}{6}
=1+2/3
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+13}{2*3}=\frac{36}{6}
=6
$