2n^2+n=300

Solution for 2n^2+n=300 equation:

2n^2+n=300

We move all terms to the left:

2n^2+n-(300)=0

a = 2; b = 1; c = -300;
Δ = b2-4ac
Δ = 12-4·2·(-300)
Δ = 2401
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2401}=49$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-49}{2*2}=\frac{-50}{4}
=-12+1/2
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+49}{2*2}=\frac{48}{4}
=12
$