n^2+19n-200=0

Solution for n^2+19n-200=0 equation:

n^2+19n-200=0

a = 1; b = 19; c = -200;
Δ = b2-4ac
Δ = 192-4·1·(-200)
Δ = 1161
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:

$\sqrt{\Delta}=\sqrt{1161}=\sqrt{9*129}=\sqrt{9}*\sqrt{129}=3\sqrt{129}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-3\sqrt{129}}{2*1}=\frac{-19-3\sqrt{129}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+3\sqrt{129}}{2*1}=\frac{-19+3\sqrt{129}}{2}
$