3j^2=225

Solution for 3j^2=225 equation:

3j^2=225

We move all terms to the left:

3j^2-(225)=0

a = 3; b = 0; c = -225;
Δ = b2-4ac
Δ = 02-4·3·(-225)
Δ = 2700
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:

$\sqrt{\Delta}=\sqrt{2700}=\sqrt{900*3}=\sqrt{900}*\sqrt{3}=30\sqrt{3}$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-30\sqrt{3}}{2*3}=\frac{0-30\sqrt{3}}{6}
=-\frac{30\sqrt{3}}{6}
=-5\sqrt{3}
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+30\sqrt{3}}{2*3}=\frac{0+30\sqrt{3}}{6}
=\frac{30\sqrt{3}}{6}
=5\sqrt{3}
$