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z2+16z=0
We add all the numbers together, and all the variables
z^2+16z=0
a = 1; b = 16; c = 0;
Δ = b2-4ac
Δ = 162-4·1·0
Δ = 256
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{256}=16$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-16}{2*1}=\frac{-32}{2} =-16 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16}{2*1}=\frac{0}{2} =0 $
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