3j^2-8j+4=0

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Solution for 3j^2-8j+4=0 equation: 



3j^2-8j+4=0

a = 3; b = -8; c = +4;
Δ = b2-4ac
Δ = -82-4·3·4
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$


$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4}{2*3}=\frac{4}{6}
=2/3
$


$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4}{2*3}=\frac{12}{6}
=2
$

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