5i(1-3i)=+i

Solution for 5i(1-3i)=+i equation:

5i(1-3i)=+i

We move all terms to the left:

5i(1-3i)-(+i)=0

We add all the numbers together, and all the variables

5i(-3i+1)-(+i)=0

We multiply parentheses

-15i^2+5i-(+i)=0

We get rid of parentheses

-15i^2+5i-i=0

We add all the numbers together, and all the variables

-15i^2+4i=0

a = -15; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·(-15)·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*-15}=\frac{-8}{-30}
=4/15
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*-15}=\frac{0}{-30}
=0
$