0=(3+7i)(3-i)

Solution for 0=(3+7i)(3-i) equation:

0=(3+7i)(3-i)

We move all terms to the left:

0-((3+7i)(3-i))=0

We add all the numbers together, and all the variables

-((7i+3)(-1i+3))+0=0

We add all the numbers together, and all the variables

-((7i+3)(-1i+3))=0

We multiply parentheses ..

-((-7i^2+21i-3i+9))=0


We calculate terms in parentheses: -((-7i^2+21i-3i+9)), so:

(-7i^2+21i-3i+9)

We get rid of parentheses

-7i^2+21i-3i+9

We add all the numbers together, and all the variables

-7i^2+18i+9

Back to the equation:

-(-7i^2+18i+9)


We get rid of parentheses

7i^2-18i-9=0

a = 7; b = -18; c = -9;
Δ = b2-4ac
Δ = -182-4·7·(-9)
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-24}{2*7}=\frac{-6}{14}
=-3/7
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+24}{2*7}=\frac{42}{14}
=3
$