(2+2i)(-1+3i)=

Solution for (2+2i)(-1+3i)= equation:

(2+2i)(-1+3i)=

We move all terms to the left:

(2+2i)(-1+3i)-()=0

We add all the numbers together, and all the variables

(2i+2)(3i-1)-()=0

We add all the numbers together, and all the variables

(2i+2)(3i-1)=0

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

6i^2+4i-2=0

a = 6; b = 4; c = -2;
Δ = b2-4ac
Δ = 42-4·6·(-2)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-8}{2*6}=\frac{-12}{12}
=-1
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+8}{2*6}=\frac{4}{12}
=1/3
$