4=3+n2

Solution for 4=3+n2 equation:

4=3+n2

We move all terms to the left:

4-(3+n2)=0

We add all the numbers together, and all the variables

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

We get rid of parentheses

-n^2-3+4=0

We add all the numbers together, and all the variables

-1n^2+1=0

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

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