-n2-16n=-12-2n2+4n

Solution for -n2-16n=-12-2n2+4n equation:

-n2-16n=-12-2n^2+4n

We move all terms to the left:

-n2-16n-(-12-2n^2+4n)=0

We add all the numbers together, and all the variables

-1n^2-(-12-2n^2+4n)-16n=0

We get rid of parentheses

-1n^2+2n^2-4n-16n+12=0

We add all the numbers together, and all the variables

n^2-20n+12=0

a = 1; b = -20; c = +12;
Δ = b2-4ac
Δ = -202-4·1·12
Δ = 352
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{352}=\sqrt{16*22}=\sqrt{16}*\sqrt{22}=4\sqrt{22}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{22}}{2*1}=\frac{20-4\sqrt{22}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{22}}{2*1}=\frac{20+4\sqrt{22}}{2}
$