0=(5+3i)(3-2i)

Solution for 0=(5+3i)(3-2i) equation:

0=(5+3i)(3-2i)

We move all terms to the left:

0-((5+3i)(3-2i))=0

We add all the numbers together, and all the variables

-((3i+5)(-2i+3))+0=0

We add all the numbers together, and all the variables

-((3i+5)(-2i+3))=0

We multiply parentheses ..

-((-6i^2+9i-10i+15))=0


We calculate terms in parentheses: -((-6i^2+9i-10i+15)), so:

(-6i^2+9i-10i+15)

We get rid of parentheses

-6i^2+9i-10i+15

We add all the numbers together, and all the variables

-6i^2-1i+15

Back to the equation:

-(-6i^2-1i+15)


We get rid of parentheses

6i^2+1i-15=0

We add all the numbers together, and all the variables

6i^2+i-15=0

a = 6; b = 1; c = -15;
Δ = b2-4ac
Δ = 12-4·6·(-15)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-19}{2*6}=\frac{-20}{12}
=-1+2/3
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+19}{2*6}=\frac{18}{12}
=1+1/2
$