(8/9)*(3i+27)=(2/9i)+68

Solution for (8/9)*(3i+27)=(2/9i)+68 equation:

(8/9)(3i+27)=(2/9i)+68

We move all terms to the left:

(8/9)(3i+27)-((2/9i)+68)=0


Domain of the equation: 9)(3i+27)!=0

i∈R



Domain of the equation: 9i)+68)!=0

i!=0/1

i!=0

i∈R


We add all the numbers together, and all the variables

(+8/9)(3i+27)-((+2/9i)+68)=0

We multiply parentheses ..

(+24i^2+8/9*27)-((+2/9i)+68)=0

We calculate fractions


(24i^2+72i)/2187i^2+()/2187i^2=0

We multiply all the terms by the denominator


(24i^2+72i)+()=0

We add all the numbers together, and all the variables

(24i^2+72i)=0

We get rid of parentheses

24i^2+72i=0

a = 24; b = 72; c = 0;
Δ = b2-4ac
Δ = 722-4·24·0
Δ = 5184
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{5184}=72$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(72)-72}{2*24}=\frac{-144}{48}
=-3
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(72)+72}{2*24}=\frac{0}{48}
=0
$