(4-i)(8-6i)=0

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Solution for (4-i)(8-6i)=0 equation:



(4-i)(8-6i)=0
We add all the numbers together, and all the variables
(-1i+4)(-6i+8)=0
We multiply parentheses ..
(+6i^2-8i-24i+32)=0
We get rid of parentheses
6i^2-8i-24i+32=0
We add all the numbers together, and all the variables
6i^2-32i+32=0
a = 6; b = -32; c = +32;
Δ = b2-4ac
Δ = -322-4·6·32
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-16}{2*6}=\frac{16}{12} =1+1/3 $
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+16}{2*6}=\frac{48}{12} =4 $

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