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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-4i^2-4i-12i-12)=0

We get rid of parentheses

-4i^2-4i-12i-12=0

We add all the numbers together, and all the variables

-4i^2-16i-12=0

a = -4; b = -16; c = -12;
Δ = b2-4ac
Δ = -162-4·(-4)·(-12)
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{-(-16)-8}{2*-4}=\frac{8}{-8}
=-1
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+8}{2*-4}=\frac{24}{-8}
=-3
$