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(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 $
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