P^2+q^2=96

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Solution for P^2+q^2=96 equation: 



^2+P^2=96

We move all terms to the left:

^2+P^2-(96)=0

determiningTheFunctionDomain
P^2-96+^2=0

We add all the numbers together, and all the variables

P^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:
$P=\frac{-b}{2a}=\frac{0}{2}=0$

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