3n(n-3)=0

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Solution for 3n(n-3)=0 equation:



3n(n-3)=0
We multiply parentheses
3n^2-9n=0
a = 3; b = -9; c = 0;
Δ = b2-4ac
Δ = -92-4·3·0
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-9}{2*3}=\frac{0}{6} =0 $
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+9}{2*3}=\frac{18}{6} =3 $

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