(5+3i)(3+5i)=0

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



(5+3i)(3+5i)=0
We add all the numbers together, and all the variables
(3i+5)(5i+3)=0
We multiply parentheses ..
(+15i^2+9i+25i+15)=0
We get rid of parentheses
15i^2+9i+25i+15=0
We add all the numbers together, and all the variables
15i^2+34i+15=0
a = 15; b = 34; c = +15;
Δ = b2-4ac
Δ = 342-4·15·15
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{-(34)-16}{2*15}=\frac{-50}{30} =-1+2/3 $
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(34)+16}{2*15}=\frac{-18}{30} =-3/5 $

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