n^2=-2+8

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

n^2=-2+8

We move all terms to the left:

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

We add all the numbers together, and all the variables

n^2-6=0

a = 1; b = 0; c = -6;
Δ = b2-4ac
Δ = 02-4·1·(-6)
Δ = 24
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:

$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{6}}{2*1}=\frac{0-2\sqrt{6}}{2}
=-\frac{2\sqrt{6}}{2}
=-\sqrt{6}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{6}}{2*1}=\frac{0+2\sqrt{6}}{2}
=\frac{2\sqrt{6}}{2}
=\sqrt{6}
$