3n^2-21=4-n^2

Solution for 3n^2-21=4-n^2 equation:

3n^2-21=4-n^2

We move all terms to the left:

3n^2-21-(4-n^2)=0

We get rid of parentheses

3n^2+n^2-4-21=0

We add all the numbers together, and all the variables

4n^2-25=0

a = 4; b = 0; c = -25;
Δ = b2-4ac
Δ = 02-4·4·(-25)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20}{2*4}=\frac{-20}{8}
=-2+1/2
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20}{2*4}=\frac{20}{8}
=2+1/2
$