(7+3i)(-5+i)=0

Solution for (7+3i)(-5+i)=0 equation:

(7+3i)(-5+i)=0

We add all the numbers together, and all the variables

(3i+7)(i-5)=0

We multiply parentheses ..

(+3i^2-15i+7i-35)=0

We get rid of parentheses

3i^2-15i+7i-35=0

We add all the numbers together, and all the variables

3i^2-8i-35=0

a = 3; b = -8; c = -35;
Δ = b2-4ac
Δ = -82-4·3·(-35)
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-22}{2*3}=\frac{-14}{6}
=-2+1/3
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+22}{2*3}=\frac{30}{6}
=5
$