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-0.5x^2+x+2=0
a = -0.5; b = 1; c = +2;
Δ = b2-4ac
Δ = 12-4·(-0.5)·2
Δ = 5
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{5}}{2*-0.5}=\frac{-1-\sqrt{5}}{-1} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{5}}{2*-0.5}=\frac{-1+\sqrt{5}}{-1} $
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