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b^2-13b+10=0
a = 1; b = -13; c = +10;
Δ = b2-4ac
Δ = -132-4·1·10
Δ = 129
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-\sqrt{129}}{2*1}=\frac{13-\sqrt{129}}{2} $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{129}}{2*1}=\frac{13+\sqrt{129}}{2} $
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