(7+3i)(7+3i)=58

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Solution for (7+3i)(7+3i)=58 equation:



(7+3i)(7+3i)=58
We move all terms to the left:
(7+3i)(7+3i)-(58)=0
We add all the numbers together, and all the variables
(3i+7)(3i+7)-58=0
We multiply parentheses ..
(+9i^2+21i+21i+49)-58=0
We get rid of parentheses
9i^2+21i+21i+49-58=0
We add all the numbers together, and all the variables
9i^2+42i-9=0
a = 9; b = 42; c = -9;
Δ = b2-4ac
Δ = 422-4·9·(-9)
Δ = 2088
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:
$\sqrt{\Delta}=\sqrt{2088}=\sqrt{36*58}=\sqrt{36}*\sqrt{58}=6\sqrt{58}$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-6\sqrt{58}}{2*9}=\frac{-42-6\sqrt{58}}{18} $
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+6\sqrt{58}}{2*9}=\frac{-42+6\sqrt{58}}{18} $

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