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n^2-3n-119=0
a = 1; b = -3; c = -119;
Δ = b2-4ac
Δ = -32-4·1·(-119)
Δ = 485
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{485}}{2*1}=\frac{3-\sqrt{485}}{2} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{485}}{2*1}=\frac{3+\sqrt{485}}{2} $
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