(4+3i)(7+-2i)=0

Solution for (4+3i)(7+-2i)=0 equation:

(4+3i)(7+-2i)=0

We add all the numbers together, and all the variables

(3i+4)(-2i)=0

We multiply parentheses ..

(-6i^2-8i)=0

We get rid of parentheses

-6i^2-8i=0

a = -6; b = -8; c = 0;
Δ = b2-4ac
Δ = -82-4·(-6)·0
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{-(-8)-8}{2*-6}=\frac{0}{-12}
=0
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8}{2*-6}=\frac{16}{-12}
=-1+1/3
$