(4-7i)(6-5i)=0

Solution for (4-7i)(6-5i)=0 equation:

(4-7i)(6-5i)=0

We add all the numbers together, and all the variables

(-7i+4)(-5i+6)=0

We multiply parentheses ..

(+35i^2-42i-20i+24)=0

We get rid of parentheses

35i^2-42i-20i+24=0

We add all the numbers together, and all the variables

35i^2-62i+24=0

a = 35; b = -62; c = +24;
Δ = b2-4ac
Δ = -622-4·35·24
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{-(-62)-22}{2*35}=\frac{40}{70}
=4/7
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-62)+22}{2*35}=\frac{84}{70}
=1+1/5
$