3n^2+n=14

Solution for 3n^2+n=14 equation:

3n^2+n=14

We move all terms to the left:

3n^2+n-(14)=0

a = 3; b = 1; c = -14;
Δ = b2-4ac
Δ = 12-4·3·(-14)
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*3}=\frac{-14}{6}
=-2+1/3
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+13}{2*3}=\frac{12}{6}
=2
$