n^2+2n=40

Solution for n^2+2n=40 equation:

n^2+2n=40

We move all terms to the left:

n^2+2n-(40)=0

a = 1; b = 2; c = -40;
Δ = b2-4ac
Δ = 22-4·1·(-40)
Δ = 164
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:

$\sqrt{\Delta}=\sqrt{164}=\sqrt{4*41}=\sqrt{4}*\sqrt{41}=2\sqrt{41}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{41}}{2*1}=\frac{-2-2\sqrt{41}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{41}}{2*1}=\frac{-2+2\sqrt{41}}{2}
$