1258=n^2+n

Solution for 1258=n^2+n equation:

1258=n^2+n

We move all terms to the left:

1258-(n^2+n)=0

We get rid of parentheses

-n^2-n+1258=0

We add all the numbers together, and all the variables

-1n^2-1n+1258=0

a = -1; b = -1; c = +1258;
Δ = b2-4ac
Δ = -12-4·(-1)·1258
Δ = 5033
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{5033}}{2*-1}=\frac{1-\sqrt{5033}}{-2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{5033}}{2*-1}=\frac{1+\sqrt{5033}}{-2}
$