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x2+0.2x+0.01=0
We add all the numbers together, and all the variables
x^2+0.2x+0.01=0
a = 1; b = 0.2; c = +0.01;
Δ = b2-4ac
Δ = 0.22-4·1·0.01
Δ = 6.9388939039072E-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{6.9388939039072E-18}}{2*1}=\frac{-0.2-\sqrt{6.9388939039072E-18}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.2)+\sqrt{6.9388939039072E-18}}{2*1}=\frac{-0.2+\sqrt{6.9388939039072E-18}}{2} $
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