(3+i)(9-8i)=0

Solution for (3+i)(9-8i)=0 equation:

(3+i)(9-8i)=0

We add all the numbers together, and all the variables

(i+3)(-8i+9)=0

We multiply parentheses ..

(-8i^2+9i-24i+27)=0

We get rid of parentheses

-8i^2+9i-24i+27=0

We add all the numbers together, and all the variables

-8i^2-15i+27=0

a = -8; b = -15; c = +27;
Δ = b2-4ac
Δ = -152-4·(-8)·27
Δ = 1089
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1089}=33$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-33}{2*-8}=\frac{-18}{-16}
=1+1/8
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+33}{2*-8}=\frac{48}{-16}
=-3
$