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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-5z^2+35z+4z-28)=0

We get rid of parentheses

-5z^2+35z+4z-28=0

We add all the numbers together, and all the variables

-5z^2+39z-28=0

a = -5; b = 39; c = -28;
Δ = b2-4ac
Δ = 392-4·(-5)·(-28)
Δ = 961
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{961}=31$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(39)-31}{2*-5}=\frac{-70}{-10}
=+7
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(39)+31}{2*-5}=\frac{-8}{-10}
=4/5
$