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

Solution for (6+5i)(-3+2i)=0 equation:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+10i^2-15i+12i-18)=0

We get rid of parentheses

10i^2-15i+12i-18=0

We add all the numbers together, and all the variables

10i^2-3i-18=0

a = 10; b = -3; c = -18;
Δ = b2-4ac
Δ = -32-4·10·(-18)
Δ = 729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{729}=27$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-27}{2*10}=\frac{-24}{20}
=-1+1/5
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+27}{2*10}=\frac{30}{20}
=1+1/2
$