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(z+8)(z+4)=0
We multiply parentheses ..
(+z^2+4z+8z+32)=0
We get rid of parentheses
z^2+4z+8z+32=0
We add all the numbers together, and all the variables
z^2+12z+32=0
a = 1; b = 12; c = +32;
Δ = b2-4ac
Δ = 122-4·1·32
Δ = 16
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{16}=4$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4}{2*1}=\frac{-16}{2} =-8 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4}{2*1}=\frac{-8}{2} =-4 $
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