n^2+12n-19=0

Solution for n^2+12n-19=0 equation:

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}
$