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

Solution for (1+i)(2+3i)=0 equation:

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

We add all the numbers together, and all the variables

(i+1)(3i+2)=0

We multiply parentheses ..

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

We get rid of parentheses

3i^2+2i+3i+2=0

We add all the numbers together, and all the variables

3i^2+5i+2=0

a = 3; b = 5; c = +2;
Δ = b2-4ac
Δ = 52-4·3·2
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-1}{2*3}=\frac{-6}{6}
=-1
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+1}{2*3}=\frac{-4}{6}
=-2/3
$