3j^2-28j+9=0

Solution for 3j^2-28j+9=0 equation:

3j^2-28j+9=0

a = 3; b = -28; c = +9;
Δ = b2-4ac
Δ = -282-4·3·9
Δ = 676
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{676}=26$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-26}{2*3}=\frac{2}{6}
=1/3
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+26}{2*3}=\frac{54}{6}
=9
$