(3=2i)^2

Solution for (3=2i)^2 equation:

(3=2i)^2

We move all terms to the left:

(3-(2i)^2)=0

We get rid of parentheses

-2i^2+3=0

a = -2; b = 0; c = +3;
Δ = b2-4ac
Δ = 02-4·(-2)·3
Δ = 24
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:

$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{6}}{2*-2}=\frac{0-2\sqrt{6}}{-4}
=-\frac{2\sqrt{6}}{-4}
=-\frac{\sqrt{6}}{-2}
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{6}}{2*-2}=\frac{0+2\sqrt{6}}{-4}
=\frac{2\sqrt{6}}{-4}
=\frac{\sqrt{6}}{-2}
$