(-4-4i)(-5-3i)=0

Solution for (-4-4i)(-5-3i)=0 equation:

(-4-4i)(-5-3i)=0

We add all the numbers together, and all the variables

(-4i-4)(-3i-5)=0

We multiply parentheses ..

(+12i^2+20i+12i+20)=0

We get rid of parentheses

12i^2+20i+12i+20=0

We add all the numbers together, and all the variables

12i^2+32i+20=0

a = 12; b = 32; c = +20;
Δ = b2-4ac
Δ = 322-4·12·20
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-8}{2*12}=\frac{-40}{24}
=-1+2/3
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+8}{2*12}=\frac{-24}{24}
=-1
$