54/7i+4=3/5i

Solution for 54/7i+4=3/5i equation:

54/7i+4=3/5i

We move all terms to the left:

54/7i+4-(3/5i)=0


Domain of the equation: 7i!=0

i!=0/7

i!=0

i∈R



Domain of the equation: 5i)!=0

i!=0/1

i!=0

i∈R


We add all the numbers together, and all the variables

54/7i-(+3/5i)+4=0

We get rid of parentheses

54/7i-3/5i+4=0

We calculate fractions


270i/35i^2+(-21i)/35i^2+4=0

We multiply all the terms by the denominator


270i+(-21i)+4*35i^2=0

Wy multiply elements

140i^2+270i+(-21i)=0

We get rid of parentheses

140i^2+270i-21i=0

We add all the numbers together, and all the variables

140i^2+249i=0

a = 140; b = 249; c = 0;
Δ = b2-4ac
Δ = 2492-4·140·0
Δ = 62001
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{62001}=249$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(249)-249}{2*140}=\frac{-498}{280}
=-1+109/140
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(249)+249}{2*140}=\frac{0}{280}
=0
$