(3z-1)(4-z)=0

Solution for (3z-1)(4-z)=0 equation:

(3z-1)(4-z)=0

We add all the numbers together, and all the variables

(3z-1)(-1z+4)=0

We multiply parentheses ..

(-3z^2+12z+z-4)=0

We get rid of parentheses

-3z^2+12z+z-4=0

We add all the numbers together, and all the variables

-3z^2+13z-4=0

a = -3; b = 13; c = -4;
Δ = b2-4ac
Δ = 132-4·(-3)·(-4)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-11}{2*-3}=\frac{-24}{-6}
=+4
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+11}{2*-3}=\frac{-2}{-6}
=1/3
$