(-3-u)(3+u)=0

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Solution for (-3-u)(3+u)=0 equation: 



(-3-u)(3+u)=0

We add all the numbers together, and all the variables

(-1u-3)(u+3)=0

We multiply parentheses ..

(-1u^2-3u-3u-9)=0

We get rid of parentheses

-1u^2-3u-3u-9=0

We add all the numbers together, and all the variables

-1u^2-6u-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:
$u=\frac{-b}{2a}=\frac{6}{-2}=-3$

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