(5-2i2)2=0

Solution for (5-2i2)2=0 equation:

(5-2i^2)2=0

We multiply parentheses

-4i^2+10=0

a = -4; b = 0; c = +10;
Δ = b2-4ac
Δ = 02-4·(-4)·10
Δ = 160
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{10}}{2*-4}=\frac{0-4\sqrt{10}}{-8}
=-\frac{4\sqrt{10}}{-8}
=-\frac{\sqrt{10}}{-2}
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{10}}{2*-4}=\frac{0+4\sqrt{10}}{-8}
=\frac{4\sqrt{10}}{-8}
=\frac{\sqrt{10}}{-2}
$