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z2+5z+5=0
We add all the numbers together, and all the variables
z^2+5z+5=0
a = 1; b = 5; c = +5;
Δ = b2-4ac
Δ = 52-4·1·5
Δ = 5
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}$$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-\sqrt{5}}{2*1}=\frac{-5-\sqrt{5}}{2} $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{5}}{2*1}=\frac{-5+\sqrt{5}}{2} $
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