(N^2+4n)-n^2=4n

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Solution for (N^2+4n)-n^2=4n equation: 



(^2+4N)-N^2=4N

We move all terms to the left:

(^2+4N)-N^2-(4N)=0

We add all the numbers together, and all the variables

-1N^2-4N+(^2+4N)=0

We get rid of parentheses

-1N^2-4N+4N+^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:
$N=\frac{-b}{2a}=\frac{0}{-2}=0$

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