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

Solution for (-3-4i)(1+2i)=0 equation:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-8i^2-4i-6i-3)=0

We get rid of parentheses

-8i^2-4i-6i-3=0

We add all the numbers together, and all the variables

-8i^2-10i-3=0

a = -8; b = -10; c = -3;
Δ = b2-4ac
Δ = -102-4·(-8)·(-3)
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2}{2*-8}=\frac{8}{-16}
=-1/2
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2}{2*-8}=\frac{12}{-16}
=-3/4
$