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0.5x^2-2x+0.5=0
a = 0.5; b = -2; c = +0.5;
Δ = b2-4ac
Δ = -22-4·0.5·0.5
Δ = 3
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{3}}{2*0.5}=\frac{2-\sqrt{3}}{1} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+\sqrt{3}}{2*0.5}=\frac{2+\sqrt{3}}{1} $
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