(1-5i)(2+i)=0

Solution for (1-5i)(2+i)=0 equation:

(1-5i)(2+i)=0

We add all the numbers together, and all the variables

(-5i+1)(i+2)=0

We multiply parentheses ..

(-5i^2-10i+i+2)=0

We get rid of parentheses

-5i^2-10i+i+2=0

We add all the numbers together, and all the variables

-5i^2-9i+2=0

a = -5; b = -9; c = +2;
Δ = b2-4ac
Δ = -92-4·(-5)·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{-(-9)-11}{2*-5}=\frac{-2}{-10}
=1/5
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+11}{2*-5}=\frac{20}{-10}
=-2
$