2n^2+19=-19n-5

Solution for 2n^2+19=-19n-5 equation:

2n^2+19=-19n-5

We move all terms to the left:

2n^2+19-(-19n-5)=0

We get rid of parentheses

2n^2+19n+5+19=0

We add all the numbers together, and all the variables

2n^2+19n+24=0

a = 2; b = 19; c = +24;
Δ = b2-4ac
Δ = 192-4·2·24
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-13}{2*2}=\frac{-32}{4}
=-8
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+13}{2*2}=\frac{-6}{4}
=-1+1/2
$