n^2+n=32

Solution for n^2+n=32 equation:

n^2+n=32

We move all terms to the left:

n^2+n-(32)=0

a = 1; b = 1; c = -32;
Δ = b2-4ac
Δ = 12-4·1·(-32)
Δ = 129
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{129}}{2*1}=\frac{-1-\sqrt{129}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{129}}{2*1}=\frac{-1+\sqrt{129}}{2}
$