n^2-3n=1258

Solution for n^2-3n=1258 equation:

n^2-3n=1258

We move all terms to the left:

n^2-3n-(1258)=0

a = 1; b = -3; c = -1258;
Δ = b2-4ac
Δ = -32-4·1·(-1258)
Δ = 5041
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{5041}=71$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-71}{2*1}=\frac{-68}{2}
=-34
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+71}{2*1}=\frac{74}{2}
=37
$