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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-4z^2+23z+35=0

a = -4; b = 23; c = +35;
Δ = b2-4ac
Δ = 232-4·(-4)·35
Δ = 1089
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1089}=33$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-33}{2*-4}=\frac{-56}{-8}
=+7
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+33}{2*-4}=\frac{10}{-8}
=-1+1/4
$