3n^2+n=80

Solution for 3n^2+n=80 equation:

3n^2+n=80

We move all terms to the left:

3n^2+n-(80)=0

a = 3; b = 1; c = -80;
Δ = b2-4ac
Δ = 12-4·3·(-80)
Δ = 961
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{961}=31$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-31}{2*3}=\frac{-32}{6}
=-5+1/3
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+31}{2*3}=\frac{30}{6}
=5
$