6j^2+31j+14=0

Solution for 6j^2+31j+14=0 equation:

6j^2+31j+14=0

a = 6; b = 31; c = +14;
Δ = b2-4ac
Δ = 312-4·6·14
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(31)-25}{2*6}=\frac{-56}{12}
=-4+2/3
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(31)+25}{2*6}=\frac{-6}{12}
=-1/2
$