(1+2i)(6+3i)=0

Solution for (1+2i)(6+3i)=0 equation:

(1+2i)(6+3i)=0

We add all the numbers together, and all the variables

(2i+1)(3i+6)=0

We multiply parentheses ..

(+6i^2+12i+3i+6)=0

We get rid of parentheses

6i^2+12i+3i+6=0

We add all the numbers together, and all the variables

6i^2+15i+6=0

a = 6; b = 15; c = +6;
Δ = b2-4ac
Δ = 152-4·6·6
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-9}{2*6}=\frac{-24}{12}
=-2
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+9}{2*6}=\frac{-6}{12}
=-1/2
$