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x^2+0.01x-0.001=0
a = 1; b = 0.01; c = -0.001;
Δ = b2-4ac
Δ = 0.012-4·1·(-0.001)
Δ = 0.0041
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{-(0.01)-\sqrt{0.0041}}{2*1}=\frac{-0.01-\sqrt{0.0041}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.01)+\sqrt{0.0041}}{2*1}=\frac{-0.01+\sqrt{0.0041}}{2} $
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