3j^2-2j-7=0

Solution for 3j^2-2j-7=0 equation:

3j^2-2j-7=0

a = 3; b = -2; c = -7;
Δ = b2-4ac
Δ = -22-4·3·(-7)
Δ = 88
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{88}=\sqrt{4*22}=\sqrt{4}*\sqrt{22}=2\sqrt{22}$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{22}}{2*3}=\frac{2-2\sqrt{22}}{6}
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{22}}{2*3}=\frac{2+2\sqrt{22}}{6}
$