4e(1-2e)=-e(3e+4)

Solution for 4e(1-2e)=-e(3e+4) equation:

4e(1-2e)=-e(3e+4)

We move all terms to the left:

4e(1-2e)-(-e(3e+4))=0

We add all the numbers together, and all the variables

4e(-2e+1)-(-e(3e+4))=0

We multiply parentheses

-8e^2+4e-(-e(3e+4))=0


We calculate terms in parentheses: -(-e(3e+4)), so:

-e(3e+4)

We multiply parentheses

-3e^2-4e

Back to the equation:

-(-3e^2-4e)


We get rid of parentheses

-8e^2+3e^2+4e+4e=0

We add all the numbers together, and all the variables

-5e^2+8e=0

a = -5; b = 8; c = 0;
Δ = b2-4ac
Δ = 82-4·(-5)·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$e_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$e_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$e_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8}{2*-5}=\frac{-16}{-10}
=1+3/5
$
$e_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8}{2*-5}=\frac{0}{-10}
=0
$