(-3z+8)(4z-3)=0

Solution for (-3z+8)(4z-3)=0 equation:

(-3z+8)(4z-3)=0

We multiply parentheses ..

(-12z^2+9z+32z-24)=0

We get rid of parentheses

-12z^2+9z+32z-24=0

We add all the numbers together, and all the variables

-12z^2+41z-24=0

a = -12; b = 41; c = -24;
Δ = b2-4ac
Δ = 412-4·(-12)·(-24)
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(41)-23}{2*-12}=\frac{-64}{-24}
=2+2/3
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+23}{2*-12}=\frac{-18}{-24}
=3/4
$