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(n+3)(n+3)=4n+8
We move all terms to the left:
(n+3)(n+3)-(4n+8)=0
We get rid of parentheses
(n+3)(n+3)-4n-8=0
We multiply parentheses ..
(+n^2+3n+3n+9)-4n-8=0
We get rid of parentheses
n^2+3n+3n-4n+9-8=0
We add all the numbers together, and all the variables
n^2+2n+1=0
a = 1; b = 2; c = +1;
Δ = b2-4ac
Δ = 22-4·1·1
Δ = 0
Delta is equal to zero, so there is only one solution to the equation
Stosujemy wzór:$n=\frac{-b}{2a}=\frac{-2}{2}=-1$
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