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

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

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

We multiply parentheses ..

(+z^2+7z+5z+35)=0

We get rid of parentheses

z^2+7z+5z+35=0

We add all the numbers together, and all the variables

z^2+12z+35=0

a = 1; b = 12; c = +35;
Δ = b2-4ac
Δ = 122-4·1·35
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2}{2*1}=\frac{-14}{2}
=-7
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2}{2*1}=\frac{-10}{2}
=-5
$