(4=2i)(5+3i)

Solution for (4=2i)(5+3i) equation:

(4=2i)(5+3i)

We move all terms to the left:

(4-(2i)(5+3i))=0

We add all the numbers together, and all the variables

(4-2i(3i+5))=0


We calculate terms in parentheses: +(4-2i(3i+5)), so:

4-2i(3i+5)

determiningTheFunctionDomain
-2i(3i+5)+4

We multiply parentheses

-6i^2-10i+4

Back to the equation:

+(-6i^2-10i+4)


We get rid of parentheses

-6i^2-10i+4=0

a = -6; b = -10; c = +4;
Δ = b2-4ac
Δ = -102-4·(-6)·4
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-14}{2*-6}=\frac{-4}{-12}
=1/3
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+14}{2*-6}=\frac{24}{-12}
=-2
$