1r^2+8r+16=0

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Solution for 1r^2+8r+16=0 equation: 



1r^2+8r+16=0

We add all the numbers together, and all the variables

r^2+8r+16=0

a = 1; b = 8; c = +16;
Δ = b2-4ac
Δ = 82-4·1·16
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$r=\frac{-b}{2a}=\frac{-8}{2}=-4$

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