(1+5i)*(-3-i)=0

Solution for (1+5i)*(-3-i)=0 equation:

(1+5i)(-3-i)=0

We add all the numbers together, and all the variables

(5i+1)(-1i-3)=0

We multiply parentheses ..

(-5i^2-15i-1i-3)=0

We get rid of parentheses

-5i^2-15i-1i-3=0

We add all the numbers together, and all the variables

-5i^2-16i-3=0

a = -5; b = -16; c = -3;
Δ = b2-4ac
Δ = -162-4·(-5)·(-3)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-14}{2*-5}=\frac{2}{-10}
=-1/5
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+14}{2*-5}=\frac{30}{-10}
=-3
$