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

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

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

We add all the numbers together, and all the variables

(2i+3)(-1i-4)=0

We multiply parentheses ..

(-2i^2-8i-3i-12)=0

We get rid of parentheses

-2i^2-8i-3i-12=0

We add all the numbers together, and all the variables

-2i^2-11i-12=0

a = -2; b = -11; c = -12;
Δ = b2-4ac
Δ = -112-4·(-2)·(-12)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-5}{2*-2}=\frac{6}{-4}
=-1+1/2
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+5}{2*-2}=\frac{16}{-4}
=-4
$