(3-5j)(2-6j)=2+4j

Solution for (3-5j)(2-6j)=2+4j equation:

(3-5j)(2-6j)=2+4j

We move all terms to the left:

(3-5j)(2-6j)-(2+4j)=0

We add all the numbers together, and all the variables

(-5j+3)(-6j+2)-(4j+2)=0

We get rid of parentheses

(-5j+3)(-6j+2)-4j-2=0

We multiply parentheses ..

(+30j^2-10j-18j+6)-4j-2=0

We get rid of parentheses

30j^2-10j-18j-4j+6-2=0

We add all the numbers together, and all the variables

30j^2-32j+4=0

a = 30; b = -32; c = +4;
Δ = b2-4ac
Δ = -322-4·30·4
Δ = 544
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:

$\sqrt{\Delta}=\sqrt{544}=\sqrt{16*34}=\sqrt{16}*\sqrt{34}=4\sqrt{34}$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-4\sqrt{34}}{2*30}=\frac{32-4\sqrt{34}}{60}
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+4\sqrt{34}}{2*30}=\frac{32+4\sqrt{34}}{60}
$