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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

n^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
$