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X^2+0.5X=0
a = 1; b = 0.5; c = 0;
Δ = b2-4ac
Δ = 0.52-4·1·0
Δ = 0.25
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{-(0.5)-\sqrt{0.25}}{2*1}=\frac{-0.5-\sqrt{0.25}}{2} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.5)+\sqrt{0.25}}{2*1}=\frac{-0.5+\sqrt{0.25}}{2} $
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