3n^2-1=-n^2

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

3n^2-1=-n^2

We move all terms to the left:

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

We get rid of parentheses

3n^2+n^2-1=0

We add all the numbers together, and all the variables

4n^2-1=0

a = 4; b = 0; c = -1;
Δ = b2-4ac
Δ = 02-4·4·(-1)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4}{2*4}=\frac{-4}{8}
=-1/2
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4}{2*4}=\frac{4}{8}
=1/2
$