P=(m+2)(m-2)(m2+4)+16

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Solution for P=(m+2)(m-2)(m2+4)+16 equation: 



=(P+2)(P-2)(P2+4)+16

We move all terms to the left:

-((P+2)(P-2)(P2+4)+16)=0

We add all the numbers together, and all the variables

-((P+2)(P-2)(+P^2+4)+16)=0

We use the square of the difference formula

P^2+4=0

a = 1; b = 0; c = +4;
Δ = b2-4ac
Δ = 02-4·1·4
Δ = -16
Delta is less than zero, so there is no solution for the equation

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