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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

2n^2-3n+1=0

a = 2; b = -3; c = +1;
Δ = b2-4ac
Δ = -32-4·2·1
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-1}{2*2}=\frac{2}{4}
=1/2
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+1}{2*2}=\frac{4}{4}
=1
$