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y^2-13y-22=0
a = 1; b = -13; c = -22;
Δ = b2-4ac
Δ = -132-4·1·(-22)
Δ = 257
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-\sqrt{257}}{2*1}=\frac{13-\sqrt{257}}{2} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{257}}{2*1}=\frac{13+\sqrt{257}}{2} $
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