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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

(2i+3)(-1i+5)-()=0

We add all the numbers together, and all the variables

(2i+3)(-1i+5)=0

We multiply parentheses ..

(-2i^2+10i-3i+15)=0

We get rid of parentheses

-2i^2+10i-3i+15=0

We add all the numbers together, and all the variables

-2i^2+7i+15=0

a = -2; b = 7; c = +15;
Δ = b2-4ac
Δ = 72-4·(-2)·15
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-13}{2*-2}=\frac{-20}{-4}
=+5
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+13}{2*-2}=\frac{6}{-4}
=-1+1/2
$