(4z-5)(3-z)=0

Solution for (4z-5)(3-z)=0 equation:

(4z-5)(3-z)=0

We add all the numbers together, and all the variables

(4z-5)(-1z+3)=0

We multiply parentheses ..

(-4z^2+12z+5z-15)=0

We get rid of parentheses

-4z^2+12z+5z-15=0

We add all the numbers together, and all the variables

-4z^2+17z-15=0

a = -4; b = 17; c = -15;
Δ = b2-4ac
Δ = 172-4·(-4)·(-15)
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{-(17)-7}{2*-4}=\frac{-24}{-8}
=+3
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+7}{2*-4}=\frac{-10}{-8}
=1+1/4
$