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0.5x^2+4x+0.2=0
a = 0.5; b = 4; c = +0.2;
Δ = b2-4ac
Δ = 42-4·0.5·0.2
Δ = 15.6
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{-(4)-\sqrt{15.6}}{2*0.5}=\frac{-4-\sqrt{15.6}}{1} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+\sqrt{15.6}}{2*0.5}=\frac{-4+\sqrt{15.6}}{1} $
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