(-2+3i)(-2-3i)=

Solution for (-2+3i)(-2-3i)= equation:

(-2+3i)(-2-3i)=

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-9i^2-6i+6i+4)=0

We get rid of parentheses

-9i^2-6i+6i+4=0

We add all the numbers together, and all the variables

-9i^2+4=0

a = -9; b = 0; c = +4;
Δ = b2-4ac
Δ = 02-4·(-9)·4
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12}{2*-9}=\frac{-12}{-18}
=2/3
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12}{2*-9}=\frac{12}{-18}
=-2/3
$