3j^2-14j+15=0

Solution for 3j^2-14j+15=0 equation:

3j^2-14j+15=0

a = 3; b = -14; c = +15;
Δ = b2-4ac
Δ = -142-4·3·15
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-4}{2*3}=\frac{10}{6}
=1+2/3
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+4}{2*3}=\frac{18}{6}
=3
$