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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses

1i^2-8i-2i+6=0

We add all the numbers together, and all the variables

i^2-10i+6=0

a = 1; b = -10; c = +6;
Δ = b2-4ac
Δ = -102-4·1·6
Δ = 76
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:

$\sqrt{\Delta}=\sqrt{76}=\sqrt{4*19}=\sqrt{4}*\sqrt{19}=2\sqrt{19}$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{19}}{2*1}=\frac{10-2\sqrt{19}}{2}
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{19}}{2*1}=\frac{10+2\sqrt{19}}{2}
$