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0.01x^2-0.1x-0.3=0
a = 0.01; b = -0.1; c = -0.3;
Δ = b2-4ac
Δ = -0.12-4·0.01·(-0.3)
Δ = 0.022
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{-(-0.1)-\sqrt{0.022}}{2*0.01}=\frac{0.1-\sqrt{0.022}}{0.02} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-0.1)+\sqrt{0.022}}{2*0.01}=\frac{0.1+\sqrt{0.022}}{0.02} $
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