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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-1i^2-3i+2i+6)=0

We get rid of parentheses

-1i^2-3i+2i+6=0

We add all the numbers together, and all the variables

-1i^2-1i+6=0

a = -1; b = -1; c = +6;
Δ = b2-4ac
Δ = -12-4·(-1)·6
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-5}{2*-1}=\frac{-4}{-2}
=+2
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+5}{2*-1}=\frac{6}{-2}
=-3
$