0=-24-22n-3n^2

Solution for 0=-24-22n-3n^2 equation:

0=-24-22n-3n^2

We move all terms to the left:

0-(-24-22n-3n^2)=0

We add all the numbers together, and all the variables

-(-24-22n-3n^2)=0

We get rid of parentheses

3n^2+22n+24=0

a = 3; b = 22; c = +24;
Δ = b2-4ac
Δ = 222-4·3·24
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-14}{2*3}=\frac{-36}{6}
=-6
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+14}{2*3}=\frac{-8}{6}
=-1+1/3
$