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0.1x^2-x+1=0
We add all the numbers together, and all the variables
0.1x^2-1x+1=0
a = 0.1; b = -1; c = +1;
Δ = b2-4ac
Δ = -12-4·0.1·1
Δ = 0.6
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{-(-1)-\sqrt{0.6}}{2*0.1}=\frac{1-\sqrt{0.6}}{0.2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{0.6}}{2*0.1}=\frac{1+\sqrt{0.6}}{0.2} $
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