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x^2+3.2x+2=0
a = 1; b = 3.2; c = +2;
Δ = b2-4ac
Δ = 3.22-4·1·2
Δ = 2.24
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{-(3.2)-\sqrt{2.24}}{2*1}=\frac{-3.2-\sqrt{2.24}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3.2)+\sqrt{2.24}}{2*1}=\frac{-3.2+\sqrt{2.24}}{2} $
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