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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+15i^2+30i+6i+12)=0

We get rid of parentheses

15i^2+30i+6i+12=0

We add all the numbers together, and all the variables

15i^2+36i+12=0

a = 15; b = 36; c = +12;
Δ = b2-4ac
Δ = 362-4·15·12
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-24}{2*15}=\frac{-60}{30}
=-2
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+24}{2*15}=\frac{-12}{30}
=-2/5
$