n^2+10n-108=-8

Solution for n^2+10n-108=-8 equation:

n^2+10n-108=-8

We move all terms to the left:

n^2+10n-108-(-8)=0

We add all the numbers together, and all the variables

n^2+10n-100=0

a = 1; b = 10; c = -100;
Δ = b2-4ac
Δ = 102-4·1·(-100)
Δ = 500
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:

$\sqrt{\Delta}=\sqrt{500}=\sqrt{100*5}=\sqrt{100}*\sqrt{5}=10\sqrt{5}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10\sqrt{5}}{2*1}=\frac{-10-10\sqrt{5}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10\sqrt{5}}{2*1}=\frac{-10+10\sqrt{5}}{2}
$