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n^2-5n-42=0
a = 1; b = -5; c = -42;
Δ = b2-4ac
Δ = -52-4·1·(-42)
Δ = 193
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{-(-5)-\sqrt{193}}{2*1}=\frac{5-\sqrt{193}}{2} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{193}}{2*1}=\frac{5+\sqrt{193}}{2} $
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