n^2+n=31.25

Solution for n^2+n=31.25 equation:

n^2+n=31.25

We move all terms to the left:

n^2+n-(31.25)=0

We add all the numbers together, and all the variables

n^2+n-31.25=0

a = 1; b = 1; c = -31.25;
Δ = b2-4ac
Δ = 12-4·1·(-31.25)
Δ = 126
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{126}=\sqrt{9*14}=\sqrt{9}*\sqrt{14}=3\sqrt{14}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-3\sqrt{14}}{2*1}=\frac{-1-3\sqrt{14}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+3\sqrt{14}}{2*1}=\frac{-1+3\sqrt{14}}{2}
$