0=n^2-3n-18

Solution for 0=n^2-3n-18 equation:

0=n^2-3n-18

We move all terms to the left:

0-(n^2-3n-18)=0

We add all the numbers together, and all the variables

-(n^2-3n-18)=0

We get rid of parentheses

-n^2+3n+18=0

We add all the numbers together, and all the variables

-1n^2+3n+18=0

a = -1; b = 3; c = +18;
Δ = b2-4ac
Δ = 32-4·(-1)·18
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-9}{2*-1}=\frac{-12}{-2}
=+6
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+9}{2*-1}=\frac{6}{-2}
=-3
$