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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-4z^2+16z+3z-12)=0

We get rid of parentheses

-4z^2+16z+3z-12=0

We add all the numbers together, and all the variables

-4z^2+19z-12=0

a = -4; b = 19; c = -12;
Δ = b2-4ac
Δ = 192-4·(-4)·(-12)
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{-(19)-13}{2*-4}=\frac{-32}{-8}
=+4
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+13}{2*-4}=\frac{-6}{-8}
=3/4
$