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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+6i^2+18i+4i+12)=0

We get rid of parentheses

6i^2+18i+4i+12=0

We add all the numbers together, and all the variables

6i^2+22i+12=0

a = 6; b = 22; c = +12;
Δ = b2-4ac
Δ = 222-4·6·12
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-14}{2*6}=\frac{-36}{12}
=-3
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+14}{2*6}=\frac{-8}{12}
=-2/3
$