n2+n=992

Solution for n2+n=992 equation:

n2+n=992

We move all terms to the left:

n2+n-(992)=0

We add all the numbers together, and all the variables

n^2+n-992=0

a = 1; b = 1; c = -992;
Δ = b2-4ac
Δ = 12-4·1·(-992)
Δ = 3969
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3969}=63$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-63}{2*1}=\frac{-64}{2}
=-32
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+63}{2*1}=\frac{62}{2}
=31
$