11n^2+0n+-19=0

Solution for 11n^2+0n+-19=0 equation:

11n^2+0n+-19=0

We add all the numbers together, and all the variables

11n^2+n=0

a = 11; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·11·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*11}=\frac{-2}{22}
=-1/11
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*11}=\frac{0}{22}
=0
$