n^2+2n=4

Solution for n^2+2n=4 equation:

n^2+2n=4

We move all terms to the left:

n^2+2n-(4)=0

a = 1; b = 2; c = -4;
Δ = b2-4ac
Δ = 22-4·1·(-4)
Δ = 20
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:

$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{5}}{2*1}=\frac{-2-2\sqrt{5}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{5}}{2*1}=\frac{-2+2\sqrt{5}}{2}
$