n^2+(n+2)=41

Solution for n^2+(n+2)=41 equation:

n^2+(n+2)=41

We move all terms to the left:

n^2+(n+2)-(41)=0

We get rid of parentheses

n^2+n+2-41=0

We add all the numbers together, and all the variables

n^2+n-39=0

a = 1; b = 1; c = -39;
Δ = b2-4ac
Δ = 12-4·1·(-39)
Δ = 157
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{157}}{2*1}=\frac{-1-\sqrt{157}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{157}}{2*1}=\frac{-1+\sqrt{157}}{2}
$