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0.483x^2+0.475165x-0.160855=0
a = 0.483; b = 0.475165; c = -0.160855;
Δ = b2-4ac
Δ = 0.4751652-4·0.483·(-0.160855)
Δ = 0.536553637225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0.475165)-\sqrt{0.536553637225}}{2*0.483}=\frac{-0.475165-\sqrt{0.536553637225}}{0.966} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.475165)+\sqrt{0.536553637225}}{2*0.483}=\frac{-0.475165+\sqrt{0.536553637225}}{0.966} $
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