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

Solution for i=(4-3i)(1-2i) equation:

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

-((+6i^2-3i-8i+4))+i=0


We calculate terms in parentheses: -((+6i^2-3i-8i+4)), so:

(+6i^2-3i-8i+4)

We get rid of parentheses

6i^2-3i-8i+4

We add all the numbers together, and all the variables

6i^2-11i+4

Back to the equation:

-(6i^2-11i+4)


We add all the numbers together, and all the variables

i-(6i^2-11i+4)=0

We get rid of parentheses

-6i^2+i+11i-4=0

We add all the numbers together, and all the variables

-6i^2+12i-4=0

a = -6; b = 12; c = -4;
Δ = b2-4ac
Δ = 122-4·(-6)·(-4)
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{3}}{2*-6}=\frac{-12-4\sqrt{3}}{-12}
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{3}}{2*-6}=\frac{-12+4\sqrt{3}}{-12}
$