n2+2n+4=168

Solution for n2+2n+4=168 equation:

n2+2n+4=168

We move all terms to the left:

n2+2n+4-(168)=0

We add all the numbers together, and all the variables

n^2+2n-164=0

a = 1; b = 2; c = -164;
Δ = b2-4ac
Δ = 22-4·1·(-164)
Δ = 660
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:

$\sqrt{\Delta}=\sqrt{660}=\sqrt{4*165}=\sqrt{4}*\sqrt{165}=2\sqrt{165}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{165}}{2*1}=\frac{-2-2\sqrt{165}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{165}}{2*1}=\frac{-2+2\sqrt{165}}{2}
$