(i)(3-2i)+(3i)(8-7i)=0

Solution for (i)(3-2i)+(3i)(8-7i)=0 equation:

(i)(3-2i)+(3i)(8-7i)=0

We add all the numbers together, and all the variables

i(-2i+3)+3i(-7i+8)=0

We multiply parentheses

-2i^2-21i^2+3i+24i=0

We add all the numbers together, and all the variables

-23i^2+27i=0

a = -23; b = 27; c = 0;
Δ = b2-4ac
Δ = 272-4·(-23)·0
Δ = 729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{729}=27$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-27}{2*-23}=\frac{-54}{-46}
=1+4/23
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+27}{2*-23}=\frac{0}{-46}
=0
$