(5z-7)(4-z)=0

Solution for (5z-7)(4-z)=0 equation:

(5z-7)(4-z)=0

We add all the numbers together, and all the variables

(5z-7)(-1z+4)=0

We multiply parentheses ..

(-5z^2+20z+7z-28)=0

We get rid of parentheses

-5z^2+20z+7z-28=0

We add all the numbers together, and all the variables

-5z^2+27z-28=0

a = -5; b = 27; c = -28;
Δ = b2-4ac
Δ = 272-4·(-5)·(-28)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-13}{2*-5}=\frac{-40}{-10}
=+4
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+13}{2*-5}=\frac{-14}{-10}
=1+2/5
$