N+n2+n+4n+5=72

Solution for N+n2+n+4n+5=72 equation:

+N2+N+4N+5=72

We move all terms to the left:

+N2+N+4N+5-(72)=0

We add all the numbers together, and all the variables

N^2+5N-67=0

a = 1; b = 5; c = -67;
Δ = b2-4ac
Δ = 52-4·1·(-67)
Δ = 293
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{-(5)-\sqrt{293}}{2*1}=\frac{-5-\sqrt{293}}{2}
$
$N_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{293}}{2*1}=\frac{-5+\sqrt{293}}{2}
$