Z=(2+5j)(3+4j)

Solution for Z=(2+5j)(3+4j) equation:

=(2+5Z)(3+4Z)

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

-((+20Z^2+15Z+8Z+6))=0


We calculate terms in parentheses: -((+20Z^2+15Z+8Z+6)), so:

(+20Z^2+15Z+8Z+6)

We get rid of parentheses

20Z^2+15Z+8Z+6

We add all the numbers together, and all the variables

20Z^2+23Z+6

Back to the equation:

-(20Z^2+23Z+6)


We get rid of parentheses

-20Z^2-23Z-6=0

a = -20; b = -23; c = -6;
Δ = b2-4ac
Δ = -232-4·(-20)·(-6)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$Z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-7}{2*-20}=\frac{16}{-40}
=-2/5
$
$Z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+7}{2*-20}=\frac{30}{-40}
=-3/4
$