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

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

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

We move all terms to the left:

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


Domain of the equation: 2)e)!=0

e!=0/1

e!=0

e∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We multiply all the terms by the denominator


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


We calculate terms in parentheses: +((+1-e*2)e), so:

(+1-e*2)e

We add all the numbers together, and all the variables

(-e*2+1)e

We multiply parentheses

-2e^2+e

Back to the equation:

+(-2e^2+e)


Wy multiply elements

(-2e^2+e)-16e^2=0

We get rid of parentheses

-2e^2-16e^2+e=0

We add all the numbers together, and all the variables

-18e^2+e=0

a = -18; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·(-18)·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$e_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$e_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$e_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*-18}=\frac{-2}{-36}
=1/18
$
$e_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*-18}=\frac{0}{-36}
=0
$