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0.02x^2+0.01x-2=0
a = 0.02; b = 0.01; c = -2;
Δ = b2-4ac
Δ = 0.012-4·0.02·(-2)
Δ = 0.1601
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{0.1601}}{2*0.02}=\frac{-0.01-\sqrt{0.1601}}{0.04} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.01)+\sqrt{0.1601}}{2*0.02}=\frac{-0.01+\sqrt{0.1601}}{0.04} $
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