(5-3i)(7+2i)=0

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Solution for (5-3i)(7+2i)=0 equation:



(5-3i)(7+2i)=0
We add all the numbers together, and all the variables
(-3i+5)(2i+7)=0
We multiply parentheses ..
(-6i^2-21i+10i+35)=0
We get rid of parentheses
-6i^2-21i+10i+35=0
We add all the numbers together, and all the variables
-6i^2-11i+35=0
a = -6; b = -11; c = +35;
Δ = b2-4ac
Δ = -112-4·(-6)·35
Δ = 961
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{961}=31$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-31}{2*-6}=\frac{-20}{-12} =1+2/3 $
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+31}{2*-6}=\frac{42}{-12} =-3+1/2 $

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