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z^2+0.1z-0.2=0
a = 1; b = 0.1; c = -0.2;
Δ = b2-4ac
Δ = 0.12-4·1·(-0.2)
Δ = 0.81
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{-(0.1)-\sqrt{0.81}}{2*1}=\frac{-0.1-\sqrt{0.81}}{2} $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.1)+\sqrt{0.81}}{2*1}=\frac{-0.1+\sqrt{0.81}}{2} $
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