n2+n=5

Solution for n2+n=5 equation:

n2+n=5

We move all terms to the left:

n2+n-(5)=0

We add all the numbers together, and all the variables

n^2+n-5=0

a = 1; b = 1; c = -5;
Δ = b2-4ac
Δ = 12-4·1·(-5)
Δ = 21
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{21}}{2*1}=\frac{-1-\sqrt{21}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{21}}{2*1}=\frac{-1+\sqrt{21}}{2}
$