E=(4-5i)(3+7i)

Solution for E=(4-5i)(3+7i) equation:

=(4-5E)(3+7E)

We move all terms to the left:

-((4-5E)(3+7E))=0

We add all the numbers together, and all the variables

-((-5E+4)(7E+3))=0

We multiply parentheses ..

-((-35E^2-15E+28E+12))=0


We calculate terms in parentheses: -((-35E^2-15E+28E+12)), so:

(-35E^2-15E+28E+12)

We get rid of parentheses

-35E^2-15E+28E+12

We add all the numbers together, and all the variables

-35E^2+13E+12

Back to the equation:

-(-35E^2+13E+12)


We get rid of parentheses

35E^2-13E-12=0

a = 35; b = -13; c = -12;
Δ = b2-4ac
Δ = -132-4·35·(-12)
Δ = 1849
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$E_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$E_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1849}=43$
$E_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-43}{2*35}=\frac{-30}{70}
=-3/7
$
$E_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+43}{2*35}=\frac{56}{70}
=4/5
$