(7+2i)(5-3i)=0

Solution for (7+2i)(5-3i)=0 equation:

(7+2i)(5-3i)=0

We add all the numbers together, and all the variables

(2i+7)(-3i+5)=0

We multiply parentheses ..

(-6i^2+10i-21i+35)=0

We get rid of parentheses

-6i^2+10i-21i+35=0

We add all the numbers together, and all the variables

-6i^2-11i+35=0

a = -6; b = -11; c = +35;
Δ = b2-4ac
Δ = -112-4·(-6)·35
Δ = 961
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{961}=31$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-31}{2*-6}=\frac{-20}{-12}
=1+2/3
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+31}{2*-6}=\frac{42}{-12}
=-3+1/2
$