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40x^2-13x-1=0
a = 40; b = -13; c = -1;
Δ = b2-4ac
Δ = -132-4·40·(-1)
Δ = 329
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{-(-13)-\sqrt{329}}{2*40}=\frac{13-\sqrt{329}}{80} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{329}}{2*40}=\frac{13+\sqrt{329}}{80} $
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