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

Solution for (4-i)(2+5i)=0 equation:

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

We add all the numbers together, and all the variables

(-1i+4)(5i+2)=0

We multiply parentheses ..

(-5i^2-2i+20i+8)=0

We get rid of parentheses

-5i^2-2i+20i+8=0

We add all the numbers together, and all the variables

-5i^2+18i+8=0

a = -5; b = 18; c = +8;
Δ = b2-4ac
Δ = 182-4·(-5)·8
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{-(18)-22}{2*-5}=\frac{-40}{-10}
=+4
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+22}{2*-5}=\frac{4}{-10}
=-2/5
$