0.5x^2+0.2x+0.02=0

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Solution for 0.5x^2+0.2x+0.02=0 equation: 



0.5x^2+0.2x+0.02=0

a = 0.5; b = 0.2; c = +0.02;
Δ = b2-4ac
Δ = 0.22-4·0.5·0.02
Δ = 6.93889390391E-18
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{6.93889390391E-18}}{2*0.5}=\frac{-0.2-\sqrt{6.93889390391E-18}}{1}
$


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.2)+\sqrt{6.93889390391E-18}}{2*0.5}=\frac{-0.2+\sqrt{6.93889390391E-18}}{1}
$

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