(2z-1)(4-z)=0

Solution for (2z-1)(4-z)=0 equation:

(2z-1)(4-z)=0

We add all the numbers together, and all the variables

(2z-1)(-1z+4)=0

We multiply parentheses ..

(-2z^2+8z+z-4)=0

We get rid of parentheses

-2z^2+8z+z-4=0

We add all the numbers together, and all the variables

-2z^2+9z-4=0

a = -2; b = 9; c = -4;
Δ = b2-4ac
Δ = 92-4·(-2)·(-4)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-7}{2*-2}=\frac{-16}{-4}
=+4
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+7}{2*-2}=\frac{-2}{-4}
=1/2
$