0=(2-i)(1-6i)

Solution for 0=(2-i)(1-6i) equation:

0=(2-i)(1-6i)

We move all terms to the left:

0-((2-i)(1-6i))=0

We add all the numbers together, and all the variables

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

-((+6i^2-1i-12i+2))=0


We calculate terms in parentheses: -((+6i^2-1i-12i+2)), so:

(+6i^2-1i-12i+2)

We get rid of parentheses

6i^2-1i-12i+2

We add all the numbers together, and all the variables

6i^2-13i+2

Back to the equation:

-(6i^2-13i+2)


We get rid of parentheses

-6i^2+13i-2=0

a = -6; b = 13; c = -2;
Δ = b2-4ac
Δ = 132-4·(-6)·(-2)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-11}{2*-6}=\frac{-24}{-12}
=+2
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+11}{2*-6}=\frac{-2}{-12}
=1/6
$