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z^2-0.4z-0.45=0
a = 1; b = -0.4; c = -0.45;
Δ = b2-4ac
Δ = -0.42-4·1·(-0.45)
Δ = 1.96
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{-(-0.4)-\sqrt{1.96}}{2*1}=\frac{0.4-\sqrt{1.96}}{2} $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-0.4)+\sqrt{1.96}}{2*1}=\frac{0.4+\sqrt{1.96}}{2} $
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