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x^2+0.042x-0.000824=0
a = 1; b = 0.042; c = -0.000824;
Δ = b2-4ac
Δ = 0.0422-4·1·(-0.000824)
Δ = 0.00506
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{-(0.042)-\sqrt{0.00506}}{2*1}=\frac{-0.042-\sqrt{0.00506}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.042)+\sqrt{0.00506}}{2*1}=\frac{-0.042+\sqrt{0.00506}}{2} $
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