3n2+n-310=10

Solution for 3n2+n-310=10 equation:

3n^2+n-310=10

We move all terms to the left:

3n^2+n-310-(10)=0

We add all the numbers together, and all the variables

3n^2+n-320=0

a = 3; b = 1; c = -320;
Δ = b2-4ac
Δ = 12-4·3·(-320)
Δ = 3841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{3841}}{2*3}=\frac{-1-\sqrt{3841}}{6}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{3841}}{2*3}=\frac{-1+\sqrt{3841}}{6}
$