-n^2+20=-12n

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

-n^2+20=-12n

We move all terms to the left:

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

We add all the numbers together, and all the variables

-1n^2-(-12n)+20=0

We get rid of parentheses

-1n^2+12n+20=0

a = -1; b = 12; c = +20;
Δ = b2-4ac
Δ = 122-4·(-1)·20
Δ = 224
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:

$\sqrt{\Delta}=\sqrt{224}=\sqrt{16*14}=\sqrt{16}*\sqrt{14}=4\sqrt{14}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{14}}{2*-1}=\frac{-12-4\sqrt{14}}{-2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{14}}{2*-1}=\frac{-12+4\sqrt{14}}{-2}
$