(6+2i)(4-3i)=0

Solution for (6+2i)(4-3i)=0 equation:

(6+2i)(4-3i)=0

We add all the numbers together, and all the variables

(2i+6)(-3i+4)=0

We multiply parentheses ..

(-6i^2+8i-18i+24)=0

We get rid of parentheses

-6i^2+8i-18i+24=0

We add all the numbers together, and all the variables

-6i^2-10i+24=0

a = -6; b = -10; c = +24;
Δ = b2-4ac
Δ = -102-4·(-6)·24
Δ = 676
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{676}=26$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-26}{2*-6}=\frac{-16}{-12}
=1+1/3
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+26}{2*-6}=\frac{36}{-12}
=-3
$