3j^2-3j=0

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Solution for 3j^2-3j=0 equation: 



3j^2-3j=0

a = 3; b = -3; c = 0;
Δ = b2-4ac
Δ = -32-4·3·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$


$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3}{2*3}=\frac{0}{6}
=0
$


$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3}{2*3}=\frac{6}{6}
=1
$

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