3n^2-1=-n

Solution for 3n^2-1=-n equation:

3n^2-1=-n

We move all terms to the left:

3n^2-1-(-n)=0

We add all the numbers together, and all the variables

3n^2-(-1n)-1=0

We get rid of parentheses

3n^2+1n-1=0

We add all the numbers together, and all the variables

3n^2+n-1=0

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