(3j+5)(3j-5)=0

Solution for (3j+5)(3j-5)=0 equation:

(3j+5)(3j-5)=0

We use the square of the difference formula

9j^2-25=0

a = 9; b = 0; c = -25;
Δ = b2-4ac
Δ = 02-4·9·(-25)
Δ = 900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{900}=30$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-30}{2*9}=\frac{-30}{18}
=-1+2/3
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+30}{2*9}=\frac{30}{18}
=1+2/3
$