(5i-3)(2i+1)=0

Solution for (5i-3)(2i+1)=0 equation:

(5i-3)(2i+1)=0

We multiply parentheses ..

(+10i^2+5i-6i-3)=0

We get rid of parentheses

10i^2+5i-6i-3=0

We add all the numbers together, and all the variables

10i^2-1i-3=0

a = 10; b = -1; c = -3;
Δ = b2-4ac
Δ = -12-4·10·(-3)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-11}{2*10}=\frac{-10}{20}
=-1/2
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+11}{2*10}=\frac{12}{20}
=3/5
$