n2+n-128=0

Solution for n2+n-128=0 equation:

n2+n-128=0

We add all the numbers together, and all the variables

n^2+n-128=0

a = 1; b = 1; c = -128;
Δ = b2-4ac
Δ = 12-4·1·(-128)
Δ = 513
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:

$\sqrt{\Delta}=\sqrt{513}=\sqrt{9*57}=\sqrt{9}*\sqrt{57}=3\sqrt{57}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-3\sqrt{57}}{2*1}=\frac{-1-3\sqrt{57}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+3\sqrt{57}}{2*1}=\frac{-1+3\sqrt{57}}{2}
$