n^2+12n=2

Solution for n^2+12n=2 equation:

n^2+12n=2

We move all terms to the left:

n^2+12n-(2)=0

a = 1; b = 12; c = -2;
Δ = b2-4ac
Δ = 122-4·1·(-2)
Δ = 152
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:

$\sqrt{\Delta}=\sqrt{152}=\sqrt{4*38}=\sqrt{4}*\sqrt{38}=2\sqrt{38}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{38}}{2*1}=\frac{-12-2\sqrt{38}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{38}}{2*1}=\frac{-12+2\sqrt{38}}{2}
$