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0.1x^2+0.2x=0
a = 0.1; b = 0.2; c = 0;
Δ = b2-4ac
Δ = 0.22-4·0.1·0
Δ = 0.04
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{-(0.2)-\sqrt{0.04}}{2*0.1}=\frac{-0.2-\sqrt{0.04}}{0.2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.2)+\sqrt{0.04}}{2*0.1}=\frac{-0.2+\sqrt{0.04}}{0.2} $
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