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n^2-12n-19=0
a = 1; b = -12; c = -19;
Δ = b2-4ac
Δ = -122-4·1·(-19)
Δ = 220
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{220}=\sqrt{4*55}=\sqrt{4}*\sqrt{55}=2\sqrt{55}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2\sqrt{55}}{2*1}=\frac{12-2\sqrt{55}}{2} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2\sqrt{55}}{2*1}=\frac{12+2\sqrt{55}}{2} $
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