3j^2-16j+16=0

Solution for 3j^2-16j+16=0 equation:

3j^2-16j+16=0

a = 3; b = -16; c = +16;
Δ = b2-4ac
Δ = -162-4·3·16
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-8}{2*3}=\frac{8}{6}
=1+1/3
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+8}{2*3}=\frac{24}{6}
=4
$