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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+5z^2+25z+4z+20)=0

We get rid of parentheses

5z^2+25z+4z+20=0

We add all the numbers together, and all the variables

5z^2+29z+20=0

a = 5; b = 29; c = +20;
Δ = b2-4ac
Δ = 292-4·5·20
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-21}{2*5}=\frac{-50}{10}
=-5
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+21}{2*5}=\frac{-8}{10}
=-4/5
$