(3-4i)(5+7i)=0

Solution for (3-4i)(5+7i)=0 equation:

(3-4i)(5+7i)=0

We add all the numbers together, and all the variables

(-4i+3)(7i+5)=0

We multiply parentheses ..

(-28i^2-20i+21i+15)=0

We get rid of parentheses

-28i^2-20i+21i+15=0

We add all the numbers together, and all the variables

-28i^2+i+15=0

a = -28; b = 1; c = +15;
Δ = b2-4ac
Δ = 12-4·(-28)·15
Δ = 1681
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1681}=41$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-41}{2*-28}=\frac{-42}{-56}
=3/4
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+41}{2*-28}=\frac{40}{-56}
=-5/7
$