(3-4j)(55+10j)=I

Solution for (3-4j)(55+10j)=I equation:

(3-4j)(55+10j)=

We move all terms to the left:

(3-4j)(55+10j)-()=0

We add all the numbers together, and all the variables

(-4j+3)(10j+55)-()=0

We add all the numbers together, and all the variables

(-4j+3)(10j+55)=0

We multiply parentheses ..

(-40j^2-220j+30j+165)=0

We get rid of parentheses

-40j^2-220j+30j+165=0

We add all the numbers together, and all the variables

-40j^2-190j+165=0

a = -40; b = -190; c = +165;
Δ = b2-4ac
Δ = -1902-4·(-40)·165
Δ = 62500
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{62500}=250$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-190)-250}{2*-40}=\frac{-60}{-80}
=3/4
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-190)+250}{2*-40}=\frac{440}{-80}
=-5+1/2
$