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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

4n^2-3n=0

a = 4; b = -3; c = 0;
Δ = b2-4ac
Δ = -32-4·4·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3}{2*4}=\frac{0}{8}
=0
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3}{2*4}=\frac{6}{8}
=3/4
$