(2+3i)(1-4i)=0

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Solution for (2+3i)(1-4i)=0 equation:



(2+3i)(1-4i)=0
We add all the numbers together, and all the variables
(3i+2)(-4i+1)=0
We multiply parentheses ..
(-12i^2+3i-8i+2)=0
We get rid of parentheses
-12i^2+3i-8i+2=0
We add all the numbers together, and all the variables
-12i^2-5i+2=0
a = -12; b = -5; c = +2;
Δ = b2-4ac
Δ = -52-4·(-12)·2
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-11}{2*-12}=\frac{-6}{-24} =1/4 $
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+11}{2*-12}=\frac{16}{-24} =-2/3 $

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