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0.05x^2+x-199999920=0
a = 0.05; b = 1; c = -199999920;
Δ = b2-4ac
Δ = 12-4·0.05·(-199999920)
Δ = 39999985
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{-(1)-\sqrt{39999985}}{2*0.05}=\frac{-1-\sqrt{39999985}}{0.1} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{39999985}}{2*0.05}=\frac{-1+\sqrt{39999985}}{0.1} $
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