(4z-7)(3-z)=0

Solution for (4z-7)(3-z)=0 equation:

(4z-7)(3-z)=0

We add all the numbers together, and all the variables

(4z-7)(-1z+3)=0

We multiply parentheses ..

(-4z^2+12z+7z-21)=0

We get rid of parentheses

-4z^2+12z+7z-21=0

We add all the numbers together, and all the variables

-4z^2+19z-21=0

a = -4; b = 19; c = -21;
Δ = b2-4ac
Δ = 192-4·(-4)·(-21)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-5}{2*-4}=\frac{-24}{-8}
=+3
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+5}{2*-4}=\frac{-14}{-8}
=1+3/4
$