n(n+1)+4=45

Solution for n(n+1)+4=45 equation:

n(n+1)+4=45

We move all terms to the left:

n(n+1)+4-(45)=0

We add all the numbers together, and all the variables

n(n+1)-41=0

We multiply parentheses

n^2+n-41=0

a = 1; b = 1; c = -41;
Δ = b2-4ac
Δ = 12-4·1·(-41)
Δ = 165
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{165}}{2*1}=\frac{-1-\sqrt{165}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{165}}{2*1}=\frac{-1+\sqrt{165}}{2}
$