3j2-14j+11=0

Solution for 3j2-14j+11=0 equation:

3j^2-14j+11=0

a = 3; b = -14; c = +11;
Δ = b2-4ac
Δ = -142-4·3·11
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{-(-14)-8}{2*3}=\frac{6}{6}
=1
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+8}{2*3}=\frac{22}{6}
=3+2/3
$