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x^2+2x+0.19=0
a = 1; b = 2; c = +0.19;
Δ = b2-4ac
Δ = 22-4·1·0.19
Δ = 3.24
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{3.24}}{2*1}=\frac{-2-\sqrt{3.24}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+\sqrt{3.24}}{2*1}=\frac{-2+\sqrt{3.24}}{2} $
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