-11i(-2+i)=0

Solution for -11i(-2+i)=0 equation:

-11i(-2+i)=0

We add all the numbers together, and all the variables

-11i(i-2)=0

We multiply parentheses

-11i^2+22i=0

a = -11; b = 22; c = 0;
Δ = b2-4ac
Δ = 222-4·(-11)·0
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-22}{2*-11}=\frac{-44}{-22}
=+2
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+22}{2*-11}=\frac{0}{-22}
=0
$