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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-3z^2+15z+7z-35)=0

We get rid of parentheses

-3z^2+15z+7z-35=0

We add all the numbers together, and all the variables

-3z^2+22z-35=0

a = -3; b = 22; c = -35;
Δ = b2-4ac
Δ = 222-4·(-3)·(-35)
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*-3}=\frac{-30}{-6}
=+5
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+8}{2*-3}=\frac{-14}{-6}
=2+1/3
$