n^2+18n+19=0

Solution for n^2+18n+19=0 equation:

n^2+18n+19=0

a = 1; b = 18; c = +19;
Δ = b2-4ac
Δ = 182-4·1·19
Δ = 248
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:

$\sqrt{\Delta}=\sqrt{248}=\sqrt{4*62}=\sqrt{4}*\sqrt{62}=2\sqrt{62}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{62}}{2*1}=\frac{-18-2\sqrt{62}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{62}}{2*1}=\frac{-18+2\sqrt{62}}{2}
$