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-0.1013x+0.0040x^2=0
a = 0.0040; b = -0.1013; c = 0;
Δ = b2-4ac
Δ = -0.10132-4·0.0040·0
Δ = 0.01026169
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{-(-0.1013)-\sqrt{0.01026169}}{2*0.0040}=\frac{0.1013-\sqrt{0.01026169}}{0.008} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-0.1013)+\sqrt{0.01026169}}{2*0.0040}=\frac{0.1013+\sqrt{0.01026169}}{0.008} $
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