(2+3i)(3+7i)=E

Solution for (2+3i)(3+7i)=E equation:

(2+3i)(3+7i)=

We move all terms to the left:

(2+3i)(3+7i)-()=0

We add all the numbers together, and all the variables

(3i+2)(7i+3)-()=0

We add all the numbers together, and all the variables

(3i+2)(7i+3)=0

We multiply parentheses ..

(+21i^2+9i+14i+6)=0

We get rid of parentheses

21i^2+9i+14i+6=0

We add all the numbers together, and all the variables

21i^2+23i+6=0

a = 21; b = 23; c = +6;
Δ = b2-4ac
Δ = 232-4·21·6
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-5}{2*21}=\frac{-28}{42}
=-2/3
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+5}{2*21}=\frac{-18}{42}
=-3/7
$