2n^2+12n+20=n+6

Solution for 2n^2+12n+20=n+6 equation:

2n^2+12n+20=n+6

We move all terms to the left:

2n^2+12n+20-(n+6)=0

We get rid of parentheses

2n^2+12n-n-6+20=0

We add all the numbers together, and all the variables

2n^2+11n+14=0

a = 2; b = 11; c = +14;
Δ = b2-4ac
Δ = 112-4·2·14
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-3}{2*2}=\frac{-14}{4}
=-3+1/2
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+3}{2*2}=\frac{-8}{4}
=-2
$