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z2+20z+64=0

We add all the numbers together, and all the variables

z^2+20z+64=0

a = 1; b = 20; c = +64;

Δ = b^{2}-4ac

Δ = 20^{2}-4·1·64

Δ = 144

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{144}=12$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-12}{2*1}=\frac{-32}{2} =-16 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+12}{2*1}=\frac{-8}{2} =-4 $

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