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2x^2-2x+0.186=0
a = 2; b = -2; c = +0.186;
Δ = b2-4ac
Δ = -22-4·2·0.186
Δ = 2.512
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{-(-2)-\sqrt{2.512}}{2*2}=\frac{2-\sqrt{2.512}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+\sqrt{2.512}}{2*2}=\frac{2+\sqrt{2.512}}{4} $
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