n^2+10n+19=0

Solution for n^2+10n+19=0 equation:

n^2+10n+19=0

a = 1; b = 10; c = +19;
Δ = b2-4ac
Δ = 102-4·1·19
Δ = 24
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:

$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{6}}{2*1}=\frac{-10-2\sqrt{6}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{6}}{2*1}=\frac{-10+2\sqrt{6}}{2}
$