(5z-7)(3-z)=0

Solution for (5z-7)(3-z)=0 equation:

(5z-7)(3-z)=0

We add all the numbers together, and all the variables

(5z-7)(-1z+3)=0

We multiply parentheses ..

(-5z^2+15z+7z-21)=0

We get rid of parentheses

-5z^2+15z+7z-21=0

We add all the numbers together, and all the variables

-5z^2+22z-21=0

a = -5; b = 22; c = -21;
Δ = b2-4ac
Δ = 222-4·(-5)·(-21)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-8}{2*-5}=\frac{-30}{-10}
=+3
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+8}{2*-5}=\frac{-14}{-10}
=1+2/5
$