6j^2-9j-15=0

Solution for 6j^2-9j-15=0 equation:

6j^2-9j-15=0

a = 6; b = -9; c = -15;
Δ = b2-4ac
Δ = -92-4·6·(-15)
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-21}{2*6}=\frac{-12}{12}
=-1
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+21}{2*6}=\frac{30}{12}
=2+1/2
$