3n^2-96=-2n

Solution for 3n^2-96=-2n equation:

3n^2-96=-2n

We move all terms to the left:

3n^2-96-(-2n)=0

We get rid of parentheses

3n^2+2n-96=0

a = 3; b = 2; c = -96;
Δ = b2-4ac
Δ = 22-4·3·(-96)
Δ = 1156
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1156}=34$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-34}{2*3}=\frac{-36}{6}
=-6
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+34}{2*3}=\frac{32}{6}
=5+1/3
$