0=(-3-2i)*(-3+2i)

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



0=(-3-2i)(-3+2i)
We move all terms to the left:
0-((-3-2i)(-3+2i))=0
We add all the numbers together, and all the variables
-((-2i-3)(2i-3))+0=0
We add all the numbers together, and all the variables
-((-2i-3)(2i-3))=0
We multiply parentheses ..
-((-4i^2+6i-6i+9))=0
We calculate terms in parentheses: -((-4i^2+6i-6i+9)), so:
(-4i^2+6i-6i+9)
We get rid of parentheses
-4i^2+6i-6i+9
We add all the numbers together, and all the variables
-4i^2+9
Back to the equation:
-(-4i^2+9)
We get rid of parentheses
4i^2-9=0
a = 4; b = 0; c = -9;
Δ = b2-4ac
Δ = 02-4·4·(-9)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12}{2*4}=\frac{-12}{8} =-1+1/2 $
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12}{2*4}=\frac{12}{8} =1+1/2 $

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