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-0.5x^2-2x+1=0
a = -0.5; b = -2; c = +1;
Δ = b2-4ac
Δ = -22-4·(-0.5)·1
Δ = 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{-(-2)-\sqrt{6}}{2*-0.5}=\frac{2-\sqrt{6}}{-1} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+\sqrt{6}}{2*-0.5}=\frac{2+\sqrt{6}}{-1} $
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