62(3j2)=4(1j)

Solution for 62(3j2)=4(1j) equation:

62(3j^2)=4(1j)

We move all terms to the left:

62(3j^2)-(4(1j))=0

determiningTheFunctionDomain
623j^2-41j=0

a = 623; b = -41; c = 0;
Δ = b2-4ac
Δ = -412-4·623·0
Δ = 1681
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1681}=41$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-41)-41}{2*623}=\frac{0}{1246}
=0
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-41)+41}{2*623}=\frac{82}{1246}
=41/623
$