(6+3i)(9-4i)=

Solution for (6+3i)(9-4i)= equation:

(6+3i)(9-4i)=

We move all terms to the left:

(6+3i)(9-4i)-()=0

We add all the numbers together, and all the variables

(3i+6)(-4i+9)-()=0

We add all the numbers together, and all the variables

(3i+6)(-4i+9)=0

We multiply parentheses ..

(-12i^2+27i-24i+54)=0

We get rid of parentheses

-12i^2+27i-24i+54=0

We add all the numbers together, and all the variables

-12i^2+3i+54=0

a = -12; b = 3; c = +54;
Δ = b2-4ac
Δ = 32-4·(-12)·54
Δ = 2601
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2601}=51$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-51}{2*-12}=\frac{-54}{-24}
=2+1/4
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+51}{2*-12}=\frac{48}{-24}
=-2
$