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x^2-5x+0.1=0
a = 1; b = -5; c = +0.1;
Δ = b2-4ac
Δ = -52-4·1·0.1
Δ = 24.6
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{-(-5)-\sqrt{24.6}}{2*1}=\frac{5-\sqrt{24.6}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{24.6}}{2*1}=\frac{5+\sqrt{24.6}}{2} $
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