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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+8i^2-4i+20i-10)=0

We get rid of parentheses

8i^2-4i+20i-10=0

We add all the numbers together, and all the variables

8i^2+16i-10=0

a = 8; b = 16; c = -10;
Δ = b2-4ac
Δ = 162-4·8·(-10)
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-24}{2*8}=\frac{-40}{16}
=-2+1/2
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+24}{2*8}=\frac{8}{16}
=1/2
$