(6+2i)(4-3i)=30i-10i

Solution for (6+2i)(4-3i)=30i-10i equation:

(6+2i)(4-3i)=30i-10i

We move all terms to the left:

(6+2i)(4-3i)-(30i-10i)=0

We add all the numbers together, and all the variables

(2i+6)(-3i+4)-(+20i)=0

We get rid of parentheses

(2i+6)(-3i+4)-20i=0

We multiply parentheses ..

(-6i^2+8i-18i+24)-20i=0

We get rid of parentheses

-6i^2+8i-18i-20i+24=0

We add all the numbers together, and all the variables

-6i^2-30i+24=0

a = -6; b = -30; c = +24;
Δ = b2-4ac
Δ = -302-4·(-6)·24
Δ = 1476
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1476}=\sqrt{36*41}=\sqrt{36}*\sqrt{41}=6\sqrt{41}$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-30)-6\sqrt{41}}{2*-6}=\frac{30-6\sqrt{41}}{-12}
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-30)+6\sqrt{41}}{2*-6}=\frac{30+6\sqrt{41}}{-12}
$