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m^2-0.8m=0
a = 1; b = -0.8; c = 0;
Δ = b2-4ac
Δ = -0.82-4·1·0
Δ = 0.64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-0.8)-\sqrt{0.64}}{2*1}=\frac{0.8-\sqrt{0.64}}{2} $$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-0.8)+\sqrt{0.64}}{2*1}=\frac{0.8+\sqrt{0.64}}{2} $
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