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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+4i^2+3i+24i+18)=0

We get rid of parentheses

4i^2+3i+24i+18=0

We add all the numbers together, and all the variables

4i^2+27i+18=0

a = 4; b = 27; c = +18;
Δ = b2-4ac
Δ = 272-4·4·18
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-21}{2*4}=\frac{-48}{8}
=-6
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+21}{2*4}=\frac{-6}{8}
=-3/4
$