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X^2+2.05X+0.15=0
a = 1; b = 2.05; c = +0.15;
Δ = b2-4ac
Δ = 2.052-4·1·0.15
Δ = 3.6025
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{-(2.05)-\sqrt{3.6025}}{2*1}=\frac{-2.05-\sqrt{3.6025}}{2} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2.05)+\sqrt{3.6025}}{2*1}=\frac{-2.05+\sqrt{3.6025}}{2} $
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