(1/2x)+x+(x+3)=100

Solution for (1/2x)+x+(x+3)=100 equation:

(1/2x)+x+(x+3)=100

We move all terms to the left:

(1/2x)+x+(x+3)-(100)=0


Domain of the equation: 2x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

(+1/2x)+x+(x+3)-100=0

We add all the numbers together, and all the variables

x+(+1/2x)+(x+3)-100=0

We get rid of parentheses

x+1/2x+x+3-100=0

We multiply all the terms by the denominator


x*2x+x*2x+3*2x-100*2x+1=0

Wy multiply elements

2x^2+2x^2+6x-200x+1=0

We add all the numbers together, and all the variables

4x^2-194x+1=0

a = 4; b = -194; c = +1;
Δ = b2-4ac
Δ = -1942-4·4·1
Δ = 37620
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{37620}=\sqrt{36*1045}=\sqrt{36}*\sqrt{1045}=6\sqrt{1045}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-194)-6\sqrt{1045}}{2*4}=\frac{194-6\sqrt{1045}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-194)+6\sqrt{1045}}{2*4}=\frac{194+6\sqrt{1045}}{8}
$