(7+5i)(2-8i)=0

Solution for (7+5i)(2-8i)=0 equation:

(7+5i)(2-8i)=0

We add all the numbers together, and all the variables

(5i+7)(-8i+2)=0

We multiply parentheses ..

(-40i^2+10i-56i+14)=0

We get rid of parentheses

-40i^2+10i-56i+14=0

We add all the numbers together, and all the variables

-40i^2-46i+14=0

a = -40; b = -46; c = +14;
Δ = b2-4ac
Δ = -462-4·(-40)·14
Δ = 4356
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4356}=66$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-46)-66}{2*-40}=\frac{-20}{-80}
=1/4
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-46)+66}{2*-40}=\frac{112}{-80}
=-1+2/5
$