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2x^2+8x+16=8
We move all terms to the left:
2x^2+8x+16-(8)=0
We add all the numbers together, and all the variables
2x^2+8x+8=0
a = 2; b = 8; c = +8;
Δ = b2-4ac
Δ = 82-4·2·8
Δ = 0
Delta is equal to zero, so there is only one solution to the equation
Stosujemy wzór:$x=\frac{-b}{2a}=\frac{-8}{4}=-2$
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