(2-3i)(4-5i)=

Solution for (2-3i)(4-5i)= equation:

(2-3i)(4-5i)=

We move all terms to the left:

(2-3i)(4-5i)-()=0

We add all the numbers together, and all the variables

(-3i+2)(-5i+4)-()=0

We add all the numbers together, and all the variables

(-3i+2)(-5i+4)=0

We multiply parentheses ..

(+15i^2-12i-10i+8)=0

We get rid of parentheses

15i^2-12i-10i+8=0

We add all the numbers together, and all the variables

15i^2-22i+8=0

a = 15; b = -22; c = +8;
Δ = b2-4ac
Δ = -222-4·15·8
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2}{2*15}=\frac{20}{30}
=2/3
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2}{2*15}=\frac{24}{30}
=4/5
$