(N+3)^2=n^2+3^2=n2^+9

Solution for (N+3)^2=n^2+3^2=n2^+9 equation:

(+3)^2=N^2+3^2=N2^+9

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We add all the numbers together, and all the variables

-(N^2+3^2)+9=0

We get rid of parentheses

-N^2+9-3^2=0

We add all the numbers together, and all the variables

-1N^2=0

a = -1; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·(-1)·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$N=\frac{-b}{2a}=\frac{0}{-2}=0$