n^2-64=-16n

Solution for n^2-64=-16n equation:

n^2-64=-16n

We move all terms to the left:

n^2-64-(-16n)=0

We get rid of parentheses

n^2+16n-64=0

a = 1; b = 16; c = -64;
Δ = b2-4ac
Δ = 162-4·1·(-64)
Δ = 512
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:

$\sqrt{\Delta}=\sqrt{512}=\sqrt{256*2}=\sqrt{256}*\sqrt{2}=16\sqrt{2}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-16\sqrt{2}}{2*1}=\frac{-16-16\sqrt{2}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16\sqrt{2}}{2*1}=\frac{-16+16\sqrt{2}}{2}
$