n^2-2n-128=0

Solution for n^2-2n-128=0 equation:

n^2-2n-128=0

a = 1; b = -2; c = -128;
Δ = b2-4ac
Δ = -22-4·1·(-128)
Δ = 516
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:

$\sqrt{\Delta}=\sqrt{516}=\sqrt{4*129}=\sqrt{4}*\sqrt{129}=2\sqrt{129}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{129}}{2*1}=\frac{2-2\sqrt{129}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{129}}{2*1}=\frac{2+2\sqrt{129}}{2}
$