(2+3i)(10-i)=)

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

(2+3i)(10-i)=)

We move all terms to the left:

(2+3i)(10-i)-())=0

We add all the numbers together, and all the variables

(3i+2)(-1i+10)-())=0

We add all the numbers together, and all the variables

(3i+2)(-1i+10)=0

We multiply parentheses ..

(-3i^2+30i-2i+20)=0

We get rid of parentheses

-3i^2+30i-2i+20=0

We add all the numbers together, and all the variables

-3i^2+28i+20=0

a = -3; b = 28; c = +20;
Δ = b2-4ac
Δ = 282-4·(-3)·20
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-32}{2*-3}=\frac{-60}{-6}
=+10
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+32}{2*-3}=\frac{4}{-6}
=-2/3
$