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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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


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

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-9i^2+21i-10

Back to the equation:

-(-9i^2+21i-10)


We get rid of parentheses

9i^2-21i+10=0

a = 9; b = -21; c = +10;
Δ = b2-4ac
Δ = -212-4·9·10
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-9}{2*9}=\frac{12}{18}
=2/3
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+9}{2*9}=\frac{30}{18}
=1+2/3
$