(k+3)(k+3)=0

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Solution for (k+3)(k+3)=0 equation: 



(k+3)(k+3)=0

We multiply parentheses ..

(+k^2+3k+3k+9)=0

We get rid of parentheses

k^2+3k+3k+9=0

We add all the numbers together, and all the variables

k^2+6k+9=0

a = 1; b = 6; c = +9;
Δ = b2-4ac
Δ = 62-4·1·9
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$k=\frac{-b}{2a}=\frac{-6}{2}=-3$

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