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0.5x^2+x-2.5=0
a = 0.5; b = 1; c = -2.5;
Δ = b2-4ac
Δ = 12-4·0.5·(-2.5)
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{-(1)-\sqrt{6}}{2*0.5}=\frac{-1-\sqrt{6}}{1} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{6}}{2*0.5}=\frac{-1+\sqrt{6}}{1} $
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