(-6+3i)(1+4i)=0

Solution for (-6+3i)(1+4i)=0 equation:

(-6+3i)(1+4i)=0

We add all the numbers together, and all the variables

(3i-6)(4i+1)=0

We multiply parentheses ..

(+12i^2+3i-24i-6)=0

We get rid of parentheses

12i^2+3i-24i-6=0

We add all the numbers together, and all the variables

12i^2-21i-6=0

a = 12; b = -21; c = -6;
Δ = b2-4ac
Δ = -212-4·12·(-6)
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{-(-21)-27}{2*12}=\frac{-6}{24}
=-1/4
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+27}{2*12}=\frac{48}{24}
=2
$