3j^2+35j-12=0

Solution for 3j^2+35j-12=0 equation:

3j^2+35j-12=0

a = 3; b = 35; c = -12;
Δ = b2-4ac
Δ = 352-4·3·(-12)
Δ = 1369
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1369}=37$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(35)-37}{2*3}=\frac{-72}{6}
=-12
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(35)+37}{2*3}=\frac{2}{6}
=1/3
$