2n^2+n=21

Solution for 2n^2+n=21 equation:

2n^2+n=21

We move all terms to the left:

2n^2+n-(21)=0

a = 2; b = 1; c = -21;
Δ = b2-4ac
Δ = 12-4·2·(-21)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-13}{2*2}=\frac{-14}{4}
=-3+1/2
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+13}{2*2}=\frac{12}{4}
=3
$