2n2+n=1275

Solution for 2n2+n=1275 equation:

2n^2+n=1275

We move all terms to the left:

2n^2+n-(1275)=0

a = 2; b = 1; c = -1275;
Δ = b2-4ac
Δ = 12-4·2·(-1275)
Δ = 10201
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{10201}=101$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-101}{2*2}=\frac{-102}{4}
=-25+1/2
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+101}{2*2}=\frac{100}{4}
=25
$