(3-2i)(6+4i)=0

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Solution for (3-2i)(6+4i)=0 equation:



(3-2i)(6+4i)=0
We add all the numbers together, and all the variables
(-2i+3)(4i+6)=0
We multiply parentheses ..
(-8i^2-12i+12i+18)=0
We get rid of parentheses
-8i^2-12i+12i+18=0
We add all the numbers together, and all the variables
-8i^2+18=0
a = -8; b = 0; c = +18;
Δ = b2-4ac
Δ = 02-4·(-8)·18
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-24}{2*-8}=\frac{-24}{-16} =1+1/2 $
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+24}{2*-8}=\frac{24}{-16} =-1+1/2 $

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